NEET PYQs

Hint

(a) For a spontaneous reaction, \(\Delta G<0\) i.e., \(\Delta H-T \Delta S<0\)
\(
\begin{aligned}
T &>\frac{\Delta H}{\Delta S} \\
& T>\left(\frac{35.5 \times 1000}{83.6}=424.6=425 \mathrm{~K}\right) \\
\therefore \quad T &>425 \mathrm{~K}
\end{aligned}
\)

Hint

(b) The system is in isolated state.
\(
\begin{array}{l}
\because \text { For an adiabatic process, } q=0 \\
\Delta U=q+w \\
\therefore \Delta U=w \\
=-p \Delta V \\
=-2.5 \mathrm{~atm} \times(4.5-2.5) \mathrm{L} \\
=-2.5 \times 2 \mathrm{~L}-\mathrm{atm} \\
=-5 \times 101.3 \mathrm{~J} \\
=-506.5 \mathrm{~J} \approx-505 \mathrm{~J} \\
\end{array}
\)

Hint

(b) For an ideal gas undergoing reversible expansion, when temperature changes from \(T_{i}\) to \(T_{f}\) and pressure changes from \(p_{i}\) to \(p_{f}\),
\(
\Delta S=n C_{p} \ln \frac{T_{f}}{T_{i}}+n R \ln \frac{p_{i}}{p_{f}}
\)
For an isothermal process, \(T_{i}=T_{f}\) so, \(\ln 1=0\)
\(
\therefore \quad \Delta S=n R \ln \frac{p_{i}}{p_{f}}
\)

Hint

(a, c) \(\Delta G=\Delta H-T \cdot \Delta S\)

For a spontaneous reaction \(\Delta G=-\mathrm{ve}\) (always) which is possible only if \(\Delta H<0\) and \(\Delta S>0\)
\(\therefore\) spontaneous at all temperatures.

If \(\Delta H<0\) and \(\Delta S=0\)
\(\Delta G=(-\mathrm{ve})-T(0)=-\mathrm{ve}\) at all temperatures.

Hint

(b)  This is Clausius-Clapeyron equation.

\(
\frac{\mathrm{d} \ln \mathrm{P}}{\mathrm{dT}}=\frac{\Delta \mathrm{H}_{\mathrm{V}}}{\mathrm{RT}^2}
\)

According to this equation, the rate at which the natural logarithm of the vapor pressure of a liquid changes with temperature is determined by the molar enthalpy of vaporization of the liquid, the ideal gas constant, and the temperature of the system.

Hint

(d) We can represent the formation of carbon dioxide from carbon and dioxygen by the equation:
\(
\mathrm{C}(\mathrm{s})+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)
\)

The data given in the question is:
\(\Delta_f H=-393.5 \mathrm{~kJ} / \mathrm{mol}\) that is 1 mole of \(\mathrm{CO}_2\) releases \(-393.5 \mathrm{~kJ} / \mathrm{mol}\) heat.

To find the heat released upon formation of \(35.2 \mathrm{~g}\) of \(\mathrm{CO}_2\) from carbon and oxygen gas, we have to first find the mass of one molecule of carbon dioxide.

One molecule of carbon dioxide consists of 1 atom of carbon and 2 atoms of oxygen. We can calculate the mass of one molecule of carbon dioxide by adding the masses of 1 atom of carbon and 2 atoms of oxygen.

Mass of carbon atom \(=12\) grams.
Mass of oxygen atom \(=16\) grams.
Mass of one molecule of carbon dioxide \(=12+2(16)=44\) grams.
Heat released by,
\(1 \mathrm{~mole}\) of \(\mathrm{CO}_2 \rightarrow-393.5 \mathrm{~kJ} / \mathrm{mol}\)
44 grams of \(\mathrm{CO}_2 \rightarrow-393.5 \mathrm{~kJ} / \mathrm{mol}\)
1 gram of \(\mathrm{CO}_2 \rightarrow \frac{-393.5}{44} \mathrm{~kJ} / \mathrm{mol}\)
\(35.2 \mathrm{~g}\) of \(\mathrm{CO}_2 \rightarrow \frac{-393.5}{44} \times 35.2 \mathrm{~g}=-315 \mathrm{KJ} / \mathrm{mole}\)

Note: Combustion of carbon to \(\mathrm{CO}_2\) is an exothermic reaction. It is denoted by a negative sign which indicates that the heat is released. In contrast, the positive sign denotes absorption of heat.

Hint

(b) The accumulation of the molecules at the surface rather than the bulk of the solid or liquid is called adsorption. We will see the effect on thermodynamic parameters for the adsorption of a gas.
We know that during the process of adsorption, there is always a decrease in the residual forces of the surface. So, there is a decrease in the surface energy which appears as heat. Therefore, we can say that adsorption is an exothermic process.
Now, we know that in the case of spontaneous adsorption of a gas, the free energy \((\Delta G)\) of the reaction can be given as
\(
\Delta G=\Delta H-T \Delta S
\)

Now, for a spontaneous reaction, the change in free energy \((\Delta G)\) needs to be negative.
Here, free energy change can be negative only if \(\Delta H\) is sufficiently negative which we can see in the equation.
Also, when a gas is adsorbed, its entropy decreases as the freedom of movement becomes restricted. So, for the adsorption of gas, \(\Delta S\) is negative.
Thus, we can conclude that for a spontaneous adsorption of gas, \(\Delta S\) is negative and therefore \(\Delta H\) should be highly negative.

So, the correct answer is (b).
Note:

Remember that for all the spontaneous processes, the change is free energy \((\Delta G)\) is negative. For all the non-spontaneous processes, the change in free energy \((\Delta G)\) is positive.

Hint

(b) \(\Delta H=\Delta U+\Delta n_{g} R T\)
Given, \(\Delta U=2.1 \mathrm{~kcal}, \Delta n_{g}=2\),
\(
R=2 \times 10^{-3} \mathrm{~kcal}, T = 300 \mathrm{~K}
\)

\(
\therefore \quad \Delta H=2.1+2 \times 2 \times 10^{-3} \times 300=3.3 \mathrm{~kcal}
\)

Again, \(\Delta G=\Delta H-T \Delta S\)
Given, \(\Delta S=20 \times 10^{-3} \mathrm{~kcal} \mathrm{~K}^{-1}\)
On putting the values of \(\Delta H\) and \(\Delta S\) in the equation, we get
\(
\begin{aligned}
\Delta G & =3.3-300 \times 20 \times 10^{-3} \\
& =3.3-6 \times 10^3 \times 10^{-3}=-2.7 \mathrm{~kcal}
\end{aligned}
\)

Hint

(a) Hint: The change in enthalpy for a reaction is actually the difference in energies in forward and backward reaction. When these energies are equal, the difference is zero.
Complete step by step solution:
The term \(\boldsymbol{\Delta H}\) represents enthalpy of a system, in chemistry. Enthalpy is the additional product of the internal energy of a system, along with the system’s volume and the pressure too. It is a state function and is used in a lot of measurements.
It shows the capacity of a system to do non-mechanical work, or radiate heat energy. In chemistry, change of enthalpy is calculated instead of the absolute enthalpy, because it is not feasible for us to know the initial point, rather the origin.
There is a condition while measuring the difference in enthalpy and that is, pressure should remain constant during the whole process.
It is to be noted that a chemical reaction that has equal energies of activation for both the forward as well as the reverse reaction has no difference of enthalpy. As the enthalpies are the same, there is no difference between them, so we can say that \(\boldsymbol{\Delta H}=0\) for the reaction.
Which gives our answer as option (a)
Note: Enthalpy itself is a very big topic. There are various types of enthalpies like enthalpy of atomisation, enthalpy of fusion, enthalpy of formation, enthalpy of adsorption, enthalpy of hydration, etc. In all these, the differences in enthalpies are taken from final state to the initial state. Enthalpy can never become a path function as it does not depend on the pathway of the reaction.

Hint

(b) Step 1: Calculating the volume of methane and propane in the mixture
The balanced equation of combustion can be written as:
\(
\begin{array}{l}
\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\
\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{l})
\end{array}
\)

Let us assume the volume of methane \(\left(\mathrm{CH}_4\right)\) in \(5 \mathrm{~L}\) mixture \(=x \mathrm{L}\)
Now, the volume of propane \(\left(\mathrm{C}_3 \mathrm{H}_8\right)\) in mixture \(=5-x L\)
From stoichiometry, 1 mole of methane consumes 2 moles of oxygen and 1 mole of propane consumes 5 moles of oxygen.
Hence, the total number of moles of oxygen consumed \(=2 x+5(5-x)=16\)
So, \(x=3\)
Hence, the volume of methane \(=3 \mathrm{~L}\)
The volume of propane \(=2 \mathrm{~L}\)
Step 2: Calculating amount of heat evolved by combustion of mixture
\(
\begin{array}{l}
\mathrm{n}_{\mathrm{CH}_4}=\frac{3 \mathrm{~L}}{22.4 \mathrm{~L}} \\
\mathrm{n}_{\mathrm{C}_3 \mathrm{H}_8}=\frac{2 \mathrm{~L}}{22.4 \mathrm{~L}}
\end{array}
\)

Heat liberated \(=\mathrm{n}_{\mathrm{CH}_4} \times \Delta \mathrm{H}_{\mathrm{CH}_4}+\mathrm{n}_{\mathrm{C}_3 \mathrm{H}_8} \times \Delta \mathrm{H}_{\mathrm{C}_3 \mathrm{H}_8}\)
Heat liberated \(=\frac{3 \times 890+2 \times 2220}{22.4}=317 \mathrm{~kJ}\)
Hence, the correct option is (b).

Hint

(b) According to Hess’s law, equation (i) is equal to equations (ii) + (iii)

\(
\therefore x=y+z
\)

Hint

(a) \(\mathrm{C}_{\text {(graphite) }}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(\mathrm{g})}\)
\(
\Delta n_{g}=1-\frac{1}{2}=\frac{1}{2}
\)
As amount of gaseous substance is increasing in product, thus \(\Delta S\) is positive for this reaction.
And we know that \(\Delta G=\Delta H-T \Delta S\)
As \(\Delta S\) is positive, thus increase in temperature will make \(T \Delta S\) more negative and \(\Delta G\) will decrease

Hint

(c) \(\Delta H_{f}=1.435 \mathrm{~kcal} / \mathrm{mol}\)
\(
\Delta S=\frac{\Delta H_{f}}{T_{f}}=\frac{1.435 \times 10^{3}}{273}=5.26 \mathrm{~cal} / \mathrm{mol} \mathrm{K}
\)

Hint

(a) \(
\Delta_{\mathrm{vap}} H^{\circ}=40.66 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)
\(
T=100+273=373 \mathrm{~K}, \Delta E=?
\)
\(
\Delta H=\Delta E+\Delta n_g R T \Rightarrow \Delta E=\Delta H-\Delta n_g R T
\)
\(\Delta n_g=\) number of gaseous moles of products – number of gaseous moles of reactants
\(
\begin{array}{l}
\mathrm{H}_2 \mathrm{O}_{(l)} \rightleftharpoons \mathrm{H}_2 \mathrm{O}_{(g)} \\
\Delta n_g=1-0=1 \\
\begin{aligned}
\Delta E & =\Delta H-R T \\
\Delta E & =\left(40.66 \times 10^3\right)-(8.314 \times 373) \\
& =37559 \mathrm{~J} / \mathrm{mol} \text { or } 37.56 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
\end{array}
\)

Hint

(d) We know that \(\Delta G=\Delta H-T \Delta S\)
\(
\begin{aligned}
&0=\Delta H-T \Delta S \quad[\because \Delta G=0] \\
&\Delta S=\frac{\Delta H}{T}=\frac{30 \times 10^{3}}{300}=100 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}
\end{aligned}
\)

Hint

\(
\begin{array}{l}
\text { (c) Given } \\
4 \mathrm{H}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_2(\mathrm{~g}) ; \Delta \mathrm{H}=-869.6 \mathrm{~kJ} \\
\text { or } 2 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{H}(\mathrm{g}) ; \Delta \mathrm{H}=869.6 \mathrm{~kJ} \\
\mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{H}(\mathrm{g}) ; \Delta \mathrm{H}=\frac{869.6}{2}=434.8 \mathrm{~kJ}
\end{array}
\)

Hint

(c) For free expansion of an ideal gas under adiabatic condition \(q=0, \Delta T=0, w=0\)

Explanation:

For free expansion of ideal gas
\(
\begin{array}{l}
P_{\text {ext }}=0 \\
W_{p v}=0
\end{array}
\)
\(q=0\) (adiabatic process)
\(
\Delta E=q+w
\)
\(
\Delta E=0
\)
\(
\Delta E=n C_{v m} \Delta T=0
\)
\(
q=0, \Delta T=0, w=0
\)

Hint

\(
\begin{array}{ll}
\text { (b) Given } & \Delta H \\
\frac{1}{2} A \longrightarrow B & +150 \dots(i)\\
3 B \longrightarrow 2 C+D & -125 \dots(ii)\\
E+A \longrightarrow 2 D & +350 \dots(iii)
\end{array}
\)
To calculate \(\Delta H\) operate
\(
\begin{array}{l}
2 \times \text { eq. (i) }+ \text { eq. (ii) }- \text { eq. (iii) } \\
\Delta H=300-125-350=-175
\end{array}
\)

Hint

(c) Activation energy is the minimum quantity of energy which the reacting species must possess in order to undergo a specified reaction.
\(\mathrm{E}_{\mathrm{a}}\)-Activation energy
Here, energy of products is higher than that of reactants.
This graph is for an endothermic reaction.
\(\triangle \mathrm{H}\) for an endothermic reaction is positive
See figure above, when the reaction reaches point \(P\), the reactants possess enough energy for the reaction to take place.
In further reaction, some energy is use up for conversion of reactants to products.
This used up energy must be less than \(\mathrm{E}_{\mathrm{a}}\), for an endothermic reaction Since,
We find that the least \(E_a\) will be more than \(\Delta H\) for an endothermic reaction since \(E_{\text {products }}>E_{\text {reactants }}\)
\(\therefore \mathrm{E}_{\mathrm{a}}> \Delta H\)
Hence, the activation energy must be more than \(\Delta H\)

Hint

(a) Given reaction is:
\(
\frac{1}{2} X_{2}+\frac{3}{2} Y_{2} \rightleftharpoons X Y_{3}
\)
We know, \(\Delta S^{\circ}=\Sigma S_{\text {products }}^{\circ}-\Sigma S_{\text {reactants }}^{\circ}\) \(=50-(30+60)=-40 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}\)
At equilibrium \(\Delta G^{\circ}=0\)
\(
\begin{aligned}
& \Delta H^{\circ}=T \Delta S^{\circ} \\
\therefore & T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{-30 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}}{-40 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}=750 \mathrm{~K}
\end{aligned}
\)

Hint

(c) When \(K_{p}>Q\), rate of forward reaction \(>\) rate of backward reaction.
\(\therefore\) Reaction is spontaneous.
When \(\Delta G^{\circ}<R T \ln Q, \Delta G^{\circ}\) is positive, reverse reaction is feasible, thus reaction is non spontaneous.

When \(K_{p}=Q\), rate of forward reaction \(=\) rate of backward reaction.
\(\therefore\) Reaction is in equilibrium.
When \(T \Delta S>\Delta H, \Delta G\) will be negative only when \(\Delta H=+\) ve.
\(\therefore\) Reaction is spontaneous and endothermic.

Hint

(d) Hint: The work done is the product of pressure and the change in volume. The ideal gases expand spontaneously into vacuum if the external pressure will be zero.
Complete step by step answer:
The work done during the free expansion of an ideal gas is always zero when the process is irreversible. Given in the question that an ideal gas expands spontaneously into vacuum.
Work done is the product of pressure and the change in volume, the expression for work done will be:
Work done \(=-P \Delta V\)
Where, \(\mathrm{P}=\) external pressure
And \(\Delta V=\) the change in volume
Now, the ideal gas expands spontaneously in vacuum hence the external pressure will be zero
Work done \(=-(0) \times \Delta V=0\)
Hence the work done \(=0\)

Hint

(c)According to Gibb’s equation,
\(
\Delta G=\Delta H-T \Delta S
\)
when \(\Delta G=0, \Delta H=T \Delta S\)
Given, \(\Delta H=40.63 \mathrm{~kJ} \mathrm{~mol}^{-1}=40.63 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\)

\(
\begin{aligned}
& \Delta S=108.8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\
\therefore & T=\frac{\Delta H}{\Delta S}=\frac{40.63 \times 10^3}{108.8}=373.43 \mathrm{~K}
\end{aligned}
\)

Hint

(d)

\(
\mathrm{Fe}_2 \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(\mathrm{g})} \rightarrow 2 \mathrm{Fe}_{(s)}+3 \mathrm{CO}_{2(\mathrm{~g})}; \Delta H=-26.8 \mathrm{~kJ} \dots(i)
\)
\(
\mathrm{FeO}_{(s)}+\mathrm{CO}_{(g)} \rightarrow \mathrm{Fe}_{(s)}+\mathrm{CO}_{2(\mathrm{~g})}; \Delta H=-16.5 \mathrm{~kJ}; \ldots \text { (ii) }
\)
\(
\mathrm{Fe}_2 \mathrm{O}_{3(s)}+\mathrm{CO}_{(g)} \rightarrow 2 \mathrm{FeO}_{(s)}+\mathrm{CO}_{2(g)}; \Delta H=? \dots(iii)
\)
Eq. (iii) can be obtained as : \((i) – 2 \times (ii)\)
\(
=-26.8-2(-16.5)=-26.8+33.0=+6.2 \mathrm{~kJ}
\)

Hint

(b) For the reaction to be spontaneous, \(\Delta G=-\mathrm{ve}\).
Given, \(\Delta H=170 \mathrm{~kJ}=170 \times 10^{3} \mathrm{~J}\)
\(
\Delta S=170 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\)
Applying, \(\Delta G=\Delta H-T \Delta S\), the value of \(\Delta G=-\mathrm{ve}\)
only when \(T \Delta S>\Delta H\), which is possible only when \(T=1110 \mathrm{~K}\)
\(\therefore \Delta G=170 \times 10^{3}-(1110 \times 170)=-18700 \mathrm{~J}\)
Thus, reaction is spontaneous at \(T=1110 \mathrm{~K}\)

Hint

(b) For the given reaction, enthalpy of reaction can be calculated as
\(=\) B.E. \((\) reactant \()-B.E.(\) product \()\)
\(
=\left[\mathrm{B.} \mathrm{E.}_{(\mathrm{C}-\mathrm{C})}+\mathrm{B.} \mathrm{E.}_{(\mathrm{H}-\mathrm{H})}+4 \times \mathrm{B.} \mathrm{E.}_{(\mathrm{C}-\mathrm{H})}\right]-\left[\mathrm{B.} \mathrm{E.}_{(\mathrm{C}-\mathrm{C})}+6 \times \mathrm{B.} \mathrm{E.}_{(\mathrm{C}-\mathrm{H})}\right]
\)
\(
=[606.10+431.37+4 \times 410.50]-[336.49+6 \times 410.50]
\)
\(
=2679.47-2799.49=-120.02 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)

Hint

(a)\(\mathrm{H}_{2}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{HCl}\)
\(\Delta H_{\text {reaction }}=\Sigma(\mathrm{B} . \mathrm{E})_{\text {reactant }}-\Sigma(\mathrm{B} . \mathrm{E})_{\text {product }}\)
\(=\left[(\mathrm{B} . \mathrm{E})_{\mathrm{H}-\mathrm{H}}+(\mathrm{B} \cdot \mathrm{E})_{\mathrm{Cl}-\mathrm{Cl}}\right]-\left[2 \mathrm{~B} \cdot \mathrm{E}_{(\mathrm{H}-\mathrm{Cl})}\right]\)
\(=434+242-(431) \times 2\)
\(\Delta H_{\text {reaction }}=-186 \mathrm{~kJ}\)
Heat of formation is the amount of heat absorbed or evolved when one mole of substance is directly obtained from its constituent element.
Hence, enthalpy of formation of \(\mathrm{HCl}=\frac{-186}{2} \mathrm{~kJ}\) \(=-93 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Hint

(d) Gas phase reaction
\(
\begin{aligned}
&\mathrm{PCl}_{5(g)} \rightleftharpoons \mathrm{PCl}_{3(g)}+\mathrm{Cl}_{2(g)} \\
&\Delta H=\Delta E+\Delta n_{g} R T
\end{aligned}
\)
\(\Delta n_{g}=\) Change in number of moles of product and reactant species.

\(
\text { Here, } \Delta \mathrm{ng}=1+1-1=1>0
\)
Since \(\Delta n_{g}=+\mathrm{ve}\), hence \(\Delta H=+\mathrm{ve}\)
also one mole of \(\mathrm{PCl}_{5}\) is dissociated into two moles
of \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) in the same phase.
Therefore, \(\Delta S=S_{\text {product }}-S_{\text {reactant }}\)
\(\Delta S=+\) ve

Hint

(b) State functions or state variables are those which depend only on the state of the system and not on how the state was reached.
\(
\left.\begin{array}{l}
q+w=\Delta E \text { (internal energy) } \\
H-T S=G \text { (free energy) }
\end{array}\right\} \text { State functions }
\)
Path function depends on the path followed during a process as well as the end states. Work and heat are the path functions.

Hint

(d) The amount of heat absorbed or released when 1 mole of a substance is directly obtained from its constituent elements is called the heat of formation or enthalpy of formation.

Equation (i) represents neutralisation reaction, (iii) represents hydrogenation reaction and (iv) represents combustion reaction.

Note: This reaction shows the formation of \(\mathrm{H}_2 \mathrm{O}\), and the \(\mathrm{X}_2\) represents the enthalpy of formation of \(\mathrm{H}_2 \mathrm{O}\).
The enthalpy of formation is the heat evolved or absorbed when one mole of substance is formed from its constituent atoms.

Hint

(b)  \(
\mathrm{HCl} \rightarrow \frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2
\)
\(
\begin{aligned}
\Delta H & =\Sigma \text { B.E. }_{(\text {products) }}-\Sigma \text { B.E. }_{(\text {reactants })} \\
& =\frac{1}{2}\left[\text { B.E. }_{\left(\mathrm{H}_2\right)}+\text { B.E. }_{\left(\mathrm{Cl}_2\right)}\right]-\text { B.E. }_{(\mathrm{HCl})} \\
& =\frac{1}{2}(430+240)-(-90)=\frac{1}{2} \times 670+90 \\
& =335+90=425 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\)

Hint

(c) The criteria for spontaneity of a process in terms of \(\Delta G\) is as follows:

• If \(\Delta G\) is negative, the process is spontaneous.
• If \(\Delta G\) is positive, the process does not occur in the forward direction. It may occur in the backward direction.
• If \(\Delta G\) is zero, the system is in equilibrium

Hint

(b) \(\Delta H=\Delta E+\Delta n_{g} R T\)
For \(\mathrm{H}_{2(g)}+\mathrm{Br}_{2(g)} \rightarrow 2 \mathrm{HBr}_{(g)}\)
\(
\Delta n_{g}=2-(1+1)=0 \text {. i.e. } \Delta H=\Delta E
\)

Hint

(b) \(\mathrm{Br}_{2(l)}+\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{BrCl}_{(g)}\)
\(
\begin{aligned}
& \Delta H=30 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta S=105 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\
& \Delta S=\frac{\Delta H}{T} \text { i.e. } 105=\frac{30}{T} \times 1000 \\
\therefore \quad T &=\frac{30 \times 1000}{105}=285.7 \mathrm{~K}
\end{aligned}
\)

Hint

Enthalpy of hydrogenation of benzene
\(
\begin{aligned}
&=3 \times \Delta H-\text { resonance energy } \\
&=3 \times(-119.5)-(-150.4)=-358.5+150.4 \\
&=-208.1 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\)

Hint

(a) For spontaneous reaction \(\Delta H=-\mathrm{ve}, \Delta S\) \(=\) +ve

Spontaneity depends upon both critical minimum energy and maximum randomness / disorder

Note: If reaction is exothermic, therefore, \(\Delta H\) is – ve and on increasing disorder, \(\Delta S\) is \(+v e\) thus, at these condition, \(\Delta G\) is negative according to following equation.
\(
\begin{array}{l}
\Delta G=\Delta H-T \Delta S \\
\Delta G=-v e .
\end{array}
\)
Hence, for spontaneous reaction \(\Delta G\) must be \(-v e\).

Hint

(c) \(\Delta G=\Delta H-T \Delta S\)
\(\Delta G=-\) ve for spontaneous reaction
When \(\Delta S=+\mathrm{ve}, \Delta H=+\mathrm{ve}\)
and \(T \Delta S>\Delta H \Rightarrow \Delta G=-\mathrm{ve}\)

Hint

(c) MgO is the oxide of weak base and we know that heat of neutralisation of 1 eq. of strong acid with strong base is \(-57.33 \mathrm{~kJ} / \mathrm{mol}\).
\(\Rightarrow\) with weak base some heat is absorbed in dissociation of weak base.
\(\Rightarrow\) Heat of neutralisation of weak base with strong acid will be less than \(-57.33 \mathrm{~kJ} / \mathrm{mol}\)

Note: The absolute value of enthalpy of neutralization of this reaction is greater than \(-57.33 \mathrm{~kJ} \mathrm{~mol}^{-1}\).
This is due to the very high enthalpy of hydration of \(\mathrm{Mg}^{2+}\) ion. The change in enthalpy of this reaction is nearly \(-146 k J\).
Thus enthalpy of neutralization per mole of \(H^{+}\)ion is nearly \(-73 k J\), which is greater than the enthalpy of neutralization strong acid against strong base i.e. \(-57.33 k J\).

Hint

(d)

\(
\begin{array}{rrr}
\mathrm{H}-\mathrm{H}+\mathrm{Br}-\mathrm{Br} & \rightarrow 2 \mathrm{H}-\mathrm{Br} \\
433+192 & 2 \times 364 \\
=625 &  =728
\end{array}
\)
Energy absorbed               Energy released
\(
\begin{array}{l}
\Delta \mathrm{H}^{\circ}=\sum \text { bond energy (reactants) }-\sum \text { bond energy (products) } \\
=433+192-2 \times 364=-103 \mathrm{~KJ}
\end{array}
\)

Hint

(b) \(\Delta G=\Delta H-T \Delta S=-382.64-298\left(\frac{-145.6}{1000}\right)\)
\(
=-382.64+43.38=-339.3 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)

Hint

(a) \(\Delta S=R \ln \frac{V_{2}}{V_{1}}\)
Here the volume of gas increases from \(V_{1}\) to \(V_{2}\) at constant temperature \(T\).

Since \(V_{2}>V_{1}\), it is obvious that the spontaneous (irreversible) isothermal expansion of a gas is accompanied by an increase in the entropy of the system and its surroundings considered together.
\(
\Delta S_{\text {sys }}+\Delta S_{\text {surr }}>0
\)

Hint

(b) The work done during the expansion of a gas can be calculated using the formula \(\mathrm{W}=-\mathrm{P} \Delta \mathrm{V}\), where \(W\) is the work done by the system (gas), \(P\) is the external pressure, and \(\Delta V\) is the change in volume. Since the external pressure is constant, we can use this formula directly to find the answer.
In this scenario, the gas expands from \(4 \mathrm{dm}^3\) to \(6 \mathrm{dm}^3\) against a constant external pressure of \(3 \mathrm{~atm}\). The change in volume is \(6 \mathrm{dm}^3-4 \mathrm{dm}^3=2 \mathrm{dm}^3\), which is equivalent to \(2 \mathrm{~L}\) (since \(1 \mathrm{dm}^3=1 \mathrm{~L}\) ). Using the conversion factor \(1 \mathrm{~L}\) \(\mathrm{atm}=101.32 \mathrm{~J}\), we can calculate the work done as:
\(
\begin{array}{l}
W=-(3 \mathrm{~atm})(2 \mathrm{~L})=-6 \mathrm{~L} \mathrm{~atm} \\
W=-6 \mathrm{~L} \mathrm{~atm} \times 101.32 \mathrm{~J} / \mathrm{L} \mathrm{~atm}=-608 \mathrm{~J}
\end{array}
\)

Hint

\(
\begin{aligned}
&\text { (b) } \mathrm{C}_{3} \mathrm{H}_{8(g)}+5 \mathrm{O}_{2(g)} \rightarrow 3 \mathrm{CO}_{2(g)}+4 \mathrm{H}_{2} \mathrm{O}_{(l)} \\
&\Delta n_{g}=3-6=-3 \\
&\Delta H=\Delta E+P \Delta V \text { or } \Delta H-\Delta E=P \Delta V \\
&\Delta H-\Delta E=\Delta n_{g} R T=-3 R T
\end{aligned}
\)

Hint

(e) (None) \(\Delta G=-P \Delta V=\) Work done
\(
\begin{aligned}
&\Delta V=\left(\frac{12}{3.31}-\frac{12}{2.25}\right) \times 10^{-3} \mathrm{~L}=-1.71 \times 10^{-3} \mathrm{~L} \\
&\Delta G=\text { Work done }=-\left(-1.71 \times 10^{-3}\right) \times P \times 101.3 \mathrm{~J} \\
&\begin{aligned}
P &=\frac{1895}{1.71 \times 10^{-3} \times 101.3}=10.93 \times 10^{3} \mathrm{~atm} \\
&=11.07 \times 10^{8} \mathrm{~Pa}
\end{aligned}
\end{aligned}
\)

Hint

\(
\text { (d) } S=\frac{q_{r e v}}{T}=\frac{6000}{273}=21.978 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}
\)

Hint

(a) Heat of solution is defined as the amount of heat evolved or absorbed when one mole of the substance is dissolved in excess of the solvent. For hydrated salt and for salts which do not form hydrates, \(\Delta H\) is positive and for anhydrous salts, it is negative.

Note: Solution so formed will be ideal if \(\Delta \mathrm{H}_{\text {soln. }}=\Delta \mathrm{H}_1+\Delta \mathrm{H}_2+\Delta \mathrm{H}_3\)
Adding (i) to (iii)
pure solute + pure solvent \(\rightarrow\) solution;
\(
\Delta \mathrm{H}=\Delta \mathrm{H}_1+\Delta \mathrm{H}_2+\Delta \mathrm{H}_3
\)

Hint

For (c) \(\Delta H_{\text {reaction }}^{\circ}=\Delta H_f^{\circ}\left(\mathrm{XeF}_4\right)-\left[\Delta H_f^{\circ}(\mathrm{Xe})+2 \Delta H_f^{\circ}\left(\mathrm{F}_2\right)\right]\)
Enthalpies of formation of elementary substances \(\mathrm{Xe}\) and \(\mathrm{F}_2\) are taken as zero.
\(
\text { Thus, } \Delta H_{\text {reaction }}^{\circ}=\Delta H_f^{\circ}\left(\mathrm{XeF}_4\right)
\)

Hint

(b) Molar heat capacity \(=75 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)
\(
\begin{aligned}
& 18 \mathrm{~g} \text { of water }=1 \text { mole }=75 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\
& 1 \mathrm{~g} \text { of water }=\frac{75}{18} \mathrm{~J} \mathrm{~K}^{-1} \\
& Q=m \times C \times \Delta t \text { or } 1000=100 \times \frac{75}{18} \times \Delta t \\
\Rightarrow & \Delta t=\frac{10 \times 18}{75}=2.4 \mathrm{~K}
\end{aligned}
\)

Hint

(a) Entropy change \((d S)\) is given by \(d S=\frac{d q}{T}\)

\(\therefore \quad\) Unit of entropy \(=\mathrm{J} / \mathrm{K} \mathrm{mol}\) (entropy per unit mole) \(=\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\)

Hint

(a) The mathematical form of first law of thermodynamics: \(q=\Delta E+W\)

Since the system is closed and insulated, \(q=0\)
Paddle work is done on system. \(\therefore W \neq 0\).
Temperature and hence internal energy of the system increases. \(\therefore \Delta E \neq 0\)

Hint

(a) (i) \(\mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}; \Delta H_{\mathrm{i}}=-94 \mathrm{kcal} / \mathrm{mole}\)
(ii) \(2 \mathrm{H}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}_{(l)}; \Delta H_{\mathrm{ii}}=-68 \times 2 \mathrm{kcal} / \mathrm{mole}\)
(iii) \(\mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(l)}; \Delta H_{\mathrm{iii}}=-213 \mathrm{kcal} / \mathrm{mole}\)
(iv) \(\mathrm{C}_{(s)}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_{4(\mathrm{~g})}; \Delta H_{\mathrm{iv}}=\) ?
By applying Hess’s law we can compute \(\Delta H_{\mathrm{iv}}\).
\(
\begin{aligned}
\therefore \Delta H_{\mathrm{iv}}=\Delta H_{\mathrm{i}}+\Delta H_{\mathrm{ii}}-\Delta H_{\mathrm{iii}} \\
\quad=(-94-68 \times 2+213) \mathrm{kcal}=-17 \mathrm{kcal}
\end{aligned}
\)

Hint

(b) The halogen are highly electronegative elements – their non-metallic character gradually decreases from fluorine to iodine. \(\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}\). Fluorine can displace chlorine, bromine and iodine. Chlorine can displace bromine and iodine and bromine can displace iodine from their salts. Iodine cannot displace flourine, chlorine and bromine. Hence reaction (b) is not feasible.

Hint

(d) The change of entropy \(d S=\frac{d q}{T}\)
From the first law of thermodynamics,
\(
\begin{aligned}
d Q=d U+P d V=C_{V} d T+P d V \\
\Rightarrow \quad \frac{d Q}{T}=C_{V} \frac{d T}{T}+\frac{P}{T} d V \\
\Rightarrow \quad \frac{d Q}{T}=C_{V} \frac{d T}{T}+\frac{R d V}{V} \quad\left[\frac{P}{T}=\frac{R}{V}\right] \\
\therefore \quad d S=C_{V} \frac{d T}{T}+R \frac{d V}{V} \\
\Rightarrow \quad \Delta S=C_{V} \ln \frac{T_{2}}{T_{1}}+R \ln \frac{V_{2}}{V_{1}} \\
& \quad[\text { for one mole of ideal gas }] \\
\text { Here } T_{2}=T_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K} . \quad \therefore \ln \frac{T_{2}}{T_{1}}=0 \\
\therefore \quad \Delta S=R \ln \frac{V_{2}}{V_{1}}=2 \ln \frac{20}{2}=2 \ln 10=4.605 \\
\therefore \quad \Delta S=4.605 \mathrm{cal} / \mathrm{mol}-\mathrm{K} \\
\text { Entropy change for } 2 \mathrm{moles} \text { of gas } \\
\quad=2 \times 4.605 \mathrm{cal} / \mathrm{K}=9.2 \mathrm{cal} / \mathrm{K}
\end{aligned}
\)

Hint

(a) \(\Delta H_{f}^{\circ}=\Sigma H_{f \text { (product) }}^{\circ}-\Sigma H_{f}^{\circ}\) (reactant)
For the given reaction
\(
\begin{aligned}
&2 \mathrm{H}_{2} \mathrm{O}_{2(l)} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}_{(l)}+\mathrm{O}_{2(g)} \\
&\Delta H_{f}^{\circ}=2 \times \Delta H_{f}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right)-2 \times \Sigma H_{f}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}_{2}\right) \\
&\quad=2 \times-286 \mathrm{~kJ} \mathrm{~mol}^{-1}-2 \times(-188) \mathrm{kJ} \mathrm{mol}^{-1} \\
&\quad=-196 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\)

Hint

(b) \(\Delta H=\Delta E+P \Delta V\)
When \(\Delta V=0 ; w=0\).
\(\Delta H=\Delta E+0\) or \(\Delta H=\Delta E\)
As \(\Delta E=q+w, \Delta E=q\)
In the present problem, \(\Delta H=500 \mathrm{~J}\),
\(\Delta H=\Delta E=500 \mathrm{~J}, q=500 \mathrm{~J}, w=0\)

Hint

(a) \(\mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}, \Delta H_{1}=-x \mathrm{~kJ} \ldots\) (i)
\(
\mathrm{CH}_{3} \mathrm{OH}+\frac{3}{2} \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}, \Delta \mathrm{H}_{2}=-y \mathrm{~kJ} \dots(ii)
\)
Subtracting (ii) from (i), we get
\(
\begin{aligned}
&\mathrm{CH}_{4}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{OH}, \Delta H_{3}=-\text { ve } \\
&\text { i.e., }-x-(-y)=-\text { ve } \\
&y-x=-\text { ve } \\
&\text { Hence, } x>y \text {. }
\end{aligned}
\)

Hint

(d) The sign and magnitude of Gibb’s free energy is a criterion of spontaneity for a process. When \(\Delta G>0\) or +ve, it means \(G_{\text {product }}>G_{\text {reactant }}\) as \(\Delta G=G_{\text {products }}-G_{\text {reactants }}\)
the reaction will not take place spontaneously, i.e. the reaction should be spontaneous in reverse direction

\(
\begin{array}{l}
\quad \underset{+4}{\mathrm{SnO}_2} \rightarrow \underset{+2}{\mathrm{SnO~} ; \quad \Delta G>0} \\
\text { (more favourable) } \\
\end{array}
\)
\(\Delta G<0\) or \(-\mathrm{ve}\), the reaction or change occurs spontaneously.
\(\underset{+4}{\mathrm{PbO}_2} \rightarrow \underset{+2}{\mathrm{PbO}}\); \(\rightarrow \Delta G<0\)
\(\quad \quad \quad \quad \text { (more favourable) }\)

Hint

(a) For a cell reaction to be spontaneous \(\Delta G^{\circ}\) should be negative.

As \(\Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}\), so the value will be -ve only when \(E_{\text {cell }}^{\circ}\) is +ve

Hint

(a) For the reactions,
\(2 \mathrm{ZnS} \rightarrow 2 \mathrm{Zn}+\mathrm{S}_{2} ; \Delta G^{\circ}=+293 \mathrm{~J} \dots(1)\)
\(2 \mathrm{Zn}+\mathrm{O}_{2} \rightarrow 2 \mathrm{ZnO} ; \Delta G^{\circ}=-616 \mathrm{~J} \dots(2) \)
\(\mathrm{S}_{2}+2 \mathrm{O}_{2} \rightarrow 2 \mathrm{SO}_{2} ; \Delta G^{\circ}=-408 \mathrm{~J} \dots(3)\)
The \(\Delta G^{\circ}\) for the reaction,
\(2 \mathrm{ZnS}+3 \mathrm{O}_{2} \rightarrow 2 \mathrm{ZnO}+2 \mathrm{SO}_{2}\) can be obtained by adding eq. (1), (2) and (3).
So, \(\Delta G^{\circ}=293-616-408=-731 \mathrm{~J}\)

Hint

\(
\text { (a) } \Delta S=\frac{Q}{T}=\frac{2930}{300}=9.77 \mathrm{~J} / \mathrm{mol} \mathrm{K}
\)

Hint

(a) \(\Delta H=\Delta E+P \Delta V\)
also \(P V=n R T\) (ideal gas equation)
\(
\begin{aligned}
\text { or } P \Delta V=\Delta n_{g} R T \\
\Delta n_{g} =\text { change in number of gaseous moles } \\
\therefore \quad \Delta H=\Delta E+\Delta n_{g} R T \Rightarrow \Delta n_{g}=2-3=-1 \\
\Rightarrow \Delta H=\Delta E-R T
\end{aligned}
\)

Hint

(b) In endothermic reactions, energy of reactants is less than energy of products. Thus, \(E_{R}<E_{P}\).
\(
\Delta H=E_{P}-E_{R}=+\text { ve }
\)

Hint

(d) \(\mathrm{S}+\frac{3}{2} \mathrm{O}_{2} \rightarrow \mathrm{SO}_{3}+2 x \mathrm{~kcal} \dots(i)\)
\(
\mathrm{SO}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{SO}_{3}+y \mathrm{~kcal} \dots(ii)
\)
By substracting equation (ii) from (i) we get,
\(
\mathrm{S}+\mathrm{O}_{2} \rightarrow \mathrm{SO}_{2}+(2 x-y) \mathrm{~kcal}
\)
The heat of formation of \(\mathrm{SO}_{2}\) is \((2 x-y) \mathrm{~kcal} / \mathrm{mole}\)

Hint

(c) The entropy of a substance increases with increase in temperature. However at absolute zero the entropy of a perfectly crystalline substance is taken as zero, which is also called as third law of thermodynamics.

Hint

(b) Change in internal energy depends upon temperature. At constant temperature, the internal energy of the gas remains constant, so \(\Delta E=0\)

Hint

(a) \(\mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} ; \Delta H=-x \mathrm{~kJ} \dots(i)\) \(\mathrm{CO}_{(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} ; \Delta H=-\frac{y}{2} \mathrm{~kJ} \dots(ii)\)
By substracting equation (ii) from (i) we get,
\(
\begin{aligned}
&\mathrm{C}_{(s)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)} \\
&\Delta H=-x-\left(-\frac{y}{2}\right)=\frac{y-2 x}{2} {\mathrm{~kJ}}
\end{aligned}
\)

Hint

(b) This is the mathematical relation of first law of thermodynamics. Here \(\Delta U=\) change in internal energy; \(\Delta Q=\) heat absorbed by the system and \(W=\) work done by the system

Hint

(d)

\(
\begin{aligned}
&\mathrm{C}_{2} \mathrm{H}_{4}+3 \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \\
&\Delta H=\Delta H_{\text {products }}-\Delta H_{\text {reactants }} \\
&=2 \times(-394)+2 \times(-286)-(52+0)=-1412 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
\)

Hint

(b) Since a catalyst affects equally both forward and backward reactions, therefore it does not affect equilibrium constant of reaction.

Hint

(c) If more trans-2-pentene is added, then its concentration in right hand side will increase. But in order to maintain the constant \(K\), concentration of cis-2-pentene will also increase. Therefore more cis-2-pentene will be formed.

Hint

(c) For a reaction to be spontaneous, \(\Delta G\) (Gibb’s free energy change) must be negative.
\(
\Delta G=\Delta H-T \Delta S
\)
\(\Delta H=\) change in enthalpy, \(\Delta S=\) change in entropy

Hint

(c) During isothermal expansion of an ideal gas, \(\Delta T=0\).
Now we know \(H=E+P V\)
\(
\begin{array}{ll}
\because & \Delta H=\Delta E+\Delta(P V)=\Delta E+\Delta(n R T) \\
\therefore & \Delta H=\Delta E+n R \Delta T=0+0=0
\end{array}
\)

Hint

 (d)  (i) The given reaction is a combustion reaction, therefore \(\Delta H \) is less than 0 . Hence, \(\Delta H \) is negative.
(ii) Since there is increase in the number of moles, therefore \(\Delta S \) is positive
(iii) Since reaction is spontaneous, therefore \(\Delta G \)is negative.

Hint

\(
\text { (b)  } \Delta n_{g}=2-4=-2, \Delta H=\Delta E-2 R T
\)

Hint

(d) If \(n_{p}<n_{r} ; \Delta n_{g}=n_{p}-n_{r}=-\mathrm{ve}\). Hence \(\Delta H<\Delta E\)

Hint

(d) Free expansion of ideal gas
\(
\begin{aligned}
&P_{\mathrm{ex}}=0 \\
&\therefore w=-P_{\mathrm{ex}} \Delta V=0 \\
&\because \text { Adiabatic process } \Rightarrow q=0 \\
&\Delta E=q+w \text { (first law of thermodynamics) } \\
&\therefore \Delta E=0 \\
&\Delta E=n C_{v} d T \Rightarrow \Delta E=0 \\
&\text { So, } q=0, \Delta T=0, w=0
\end{aligned}
\)

Hint

\(
\text { (c) } W=-P_{\text {and }}\left(V_{f}-V\right)=-10^{5}\left(10^{-2}-10^{-3}\right)=-900 \mathrm{~J}
\)

Hint

(b)
For an ideal gas,
\(
C_{p}-C_{V}=n R
\)
(where, \(n=\) number of moles of gas,
\(C_{p^{\prime}} C_{V}=\) specific heat at constant
pressure and volume, \(R=\) universal gas constant)
As \(n=1\), so \(C_{p}-C_{V}=R\)

Hint

(c)
The change in internal energy depends on the temperature. For isothermal process, \(\Delta T=0\).
So, \(\Delta U=0\).
With an expansion of an ideal gas, more space is available for the gaseous particles.
\(\therefore\) Entropy of gas increases so, entropy of system is not zero.
i.e. \(\Delta S \neq 0\)

Hint

(d) According to Boyle’s law,
\(
\begin{array}{l}
P V=n R T \\
P=n R T\left(\frac{1}{V}\right)
\end{array}
\)

P versus \(\left(\frac{1}{V}\right)\) gives straight line graph with slope \(nRT\).

Hint

(a) Correct relation between change in enthalpy and change in internal energy is
\(
\Delta H=\Delta U+\Delta n_g R T
\)

Hint

(c) Dalton’s law of partial pressure :
Partial pressure of gas \(=\) mole fraction of gas in gaseous mixture \(\times\) Total pressure of gaseous mixture.
\(
\begin{array}{l}
\mathrm{p}_1=\mathrm{X}_1 \mathrm{p} \\
\mathrm{p}_2=\mathrm{X}_2 \mathrm{p} \\
\mathrm{p}_3=\mathrm{X}_3 \mathrm{p}
\end{array}
\)

Total pressure,
\(
\mathrm{p}=\mathrm{p}_1+\mathrm{p}_2+\mathrm{p}_3
\)

Therefore, statement-3 is incorrect.

Hint

(c)

\(
\begin{array}{ll}
\mathrm{V}=10 \mathrm{~L} & \mathrm{~W}_{\mathrm{O}_2}=64 \mathrm{~g} \\
\mathrm{~T}=27^{\circ} \mathrm{C} & \mathrm{n}_{\mathrm{O}_2}=2 \\
\mathrm{R}=0.083 . \mathrm{L} \text { bar K }^{-1} \mathrm{~mol}^{-1}
\end{array}
\)

Ideal gas equation \(\mathrm{PV}=\mathrm{nRT}\)
\(
\begin{array}{l}
P=\frac{2 \times 0.0831 \times 300}{10} \\
P=4.9 \text { bar }
\end{array}
\)

Hint

(c) Since it is isothermal, \(\Delta T =0\)
\(
\Delta U = nC _{ v } \Delta T =0
\)
Since expansion is taking place against vacuum
\(
\begin{aligned}
& P_{\text {ext }}=0 \\
& W=-P_{\text {ext }} \Delta V=0
\end{aligned}
\)
From first law of thermodynamics,
\(
\begin{aligned}
& \Delta U=q+W \\
& 0=q+0 \\
& q=0
\end{aligned}
\)

Hint

(a) For endothermic reactions, \(\Delta H_r=\) positive (Heat is absorbed)
and \(
q_p=+v e
\)

Hint

(a) To determine the internal energy change \((\Delta U )\) for the reaction at 300 K , we will use the relationship between enthalpy change \((\Delta H)\) and internal energy change:
\(
\Delta H=\Delta U+\Delta n R T
\)
where \(\Delta H\) is the enthalpy change, \(\Delta U\) is the internal energy change, \(\Delta n\) is the change in moles of gas, \(R\) is the universal gas constant, and \(T\) is the temperature.
Given:
\(
\begin{aligned}
& \Delta H=+15 kJ=15000 J \\
& T=300 K \\
& R=8.314 J mol ^{-1} K^{-1}
\end{aligned}
\)
\(\Delta n=\) moles of products – moles of reactants
From the balanced chemical equation:
\(
A _2(g)+3 B_2(g) \rightarrow 2 AB _3(g)
[latex]
Moles of reactants [latex]=1\left(\right.\) for \(\left.A _2\right)+3\left(\right.\) for \(\left.B _2\right)=4\) moles
Moles of products \(=2\left(\right.\) for \(\left.AB _3\right)\)
Therefore, \(\Delta n=2\) (products) -4 (reactants) \(=-2\)
Now, substituting these values into the equation:
\(
\begin{aligned}
& \Delta H=\Delta U+\Delta n R T \\
& 15000=\Delta U+(-2)(8.314)(300)
\end{aligned}
\)
Simplify the equation:
\(
15000=\Delta U-4988.4
\)
Therefore:
\(
\Delta U=15000+4988.4=19988.4 J
\)
Hence, the internal energy change is:
Option A: 19988.4 J

Hint

(c) A. A liquid evaporates to vapour.

During the evaporation of a liquid to form a vapour, the molecules of the substance move from a relatively ordered state (liquid) to a more disordered state (vapour). In a vapour, the molecules have more freedom of motion and are less confined than in a liquid. This transition from liquid to vapour increases the randomness or disorder of the system, hence, the entropy increases.
B. Temperature of a crystalline solid lowered from 130 K to 0 K .

Reducing the temperature of a crystalline solid generally decreases the entropy of the system. As the temperature decreases, the molecular motion within the solid becomes more restricted, leading to a decrease in randomness. At absolute zero \((0 K)\), the entropy is at its lowest possible value (ideally zero for a perfect crystal), as the molecular motion is minimized to only quantum mechanical vibrations.
c. \(2 NaHCO _{3(s)} \rightarrow Na _2 CO _{3(s)}+ CO _{2(g)}+ H _2 O _{(g)}\)

In this chemical reaction, solid sodium bicarbonate decomposes to form solid sodium carbonate, carbon dioxide gas, and water vapor. The formation of gases from a solid significantly increases the entropy of the system because gases have much higher randomness due to their free molecular motion compared to solids.
D. \(Cl _{2(g)} \rightarrow 2 Cl _{(g)}\)
The dissociation of chlorine gas \(\left( Cl _2\right)\) into atomic chlorine \(( Cl )\) represents a transition from a diatomic molecule to two separate atoms. This increases the number of particles in the gas phase, leading to increased randomness and disorder, hence, increasing the entropy.

Conclusion: Based on our analysis, processes A, C, and D involve an increase in entropy, as they each lead to greater disorder or randomness in the system. Option B is incorrect as it wrongly includes lowering the temperature of a solid as increasing entropy. Therefore, the correct answer is:

Option C :
\(A, C\) and \(D\)

Hint

(d) The question involves matching different thermodynamic processes with their corresponding description or characteristics. Let’s analyze these processes:
Isothermal process (A): This process occurs at a constant temperature. Hence, \(A\) should match with “Carried out at constant temperature” (II).
Isochoric process (B): This process occurs at a constant volume. Therefore, \(B\) should be matched with “Carried out at constant volume” (III).
Isobaric process (C): This process happens at a constant pressure, making C correspond to “Carried out at constant pressure” (IV).
Adiabatic process (D): During this process, there is no heat exchange with the surroundings. Consequently, D should be paired with “No heat exchange” (I).
The correct matches, as elaborated above, are A-II, B-III, C-IV, D-I. Therefore, the correct answer here is Option D.

Hint

(b) The work done in a reversible isothermal process can be calculated using the formula:
\(
W=-n R T \ln \left(\frac{V_f}{V_i}\right)
\)
Here:
\(W\) is the work done by the gas.
\(n\) is the number of moles of the gas.
\(R\) is the universal gas constant.
\(T\) is the temperature in Kelvin.
\(V_i\) and \(V_f\) are the initial and final volumes of the gas, respectively.
However, since the volumes are not directly provided but the pressures are given, we use the ideal gas law, \(P V=n R T\), to relate pressures and volumes at the same temperature and amount of gas:
For an ideal gas undergoing a change at constant temperature, we can also write the work done in terms of the initial and final pressures:
\(
W=-n R T \ln \left(\frac{P_i}{P_f}\right)
\)
where:
\(P_i\) and \(P_f\) are the initial and final pressures, respectively.
Given:
Substituting all values into the formula:
\(
\begin{aligned}
& W=-1 \cdot 2.0 \cdot 298.15 \ln \left(\frac{20}{10}\right) \\
& W=-2.0 \cdot 298.15 \cdot \ln (2)
\end{aligned}
\)
Now, solving this using the value of \(\ln (2) \approx 0.693\) :
\(
\begin{aligned}
& W \approx-2.0 \cdot 298.15 \cdot 0.693 \\
& W \approx-413.14 \text { calories }
\end{aligned}
\)
Thus, the work done during the process is about -413.14 calories, indicating that this amount of energy was done by the system (expansion work being done by the gas against external pressure), and hence is negative as it indicates work done by the system. Therefore, the correct option is:
\(
\begin{aligned}
& n=1 \text { (one mole of hydrogen) } \\
& T=25^{\circ} C =25+273.15=298.15 K \\
& R=2.0 cal K ^{-1} mol^{-1} \\
& P_i=20 atm \\
& P_f=10 atm
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\text { Decomposition for } 1 \text { mole of water }\\
&\begin{aligned}
& H _2 O ( g ) \rightarrow H _2(g)+\frac{1}{2} O _2(g) ; \Delta H =+\frac{483.64}{2} \\
& \Delta H =+241.82 kJ
\end{aligned}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&A + B \rightleftharpoons C + D\\
&\begin{aligned}
& {[A]=2 mol L ^{-1}} \\
& {[B]=3 mol L ^{-1}} \\
& {[C]=10 mol L ^{-1}} \\
& {[D]=6 mol L ^{-1}}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
\Delta G ^0 & =-2.303 RT \log K _{ eq } \\
& =-2.303 RT \log \frac{[ C ][ D ]}{[ A ][ B ]} \\
& =-2.303 \times 2 \times 300 \times \log \frac{10 \times 6}{2 \times 3} \\
& =-2.303 \times 2 \times 300 \times \log 10 \\
& =-1381.8 cal
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\Delta U = nC _{ V } \Delta T\\
&\text { For isothermal condition; } \Delta T=0\\
&\therefore \Delta U =0
\end{aligned}
\)

Hint

(b) At constant temperature and amount
\(
\begin{aligned}
& P_1 V_1=P_2 V_2 \\
& P_1 V_1=\frac{P_1}{3} V_2\left[\therefore P_2=\frac{P_1}{3}\right] \\
& V_2=3 V_1
\end{aligned}
\)
mole of \(O_2(g)=\frac{3.2}{32}=0.1\) mole
Volume of \(O_2(g)=(0.1 \times 22.4) L =2.24 L\)
At \(\operatorname{STP}\left( V _1\right)\)
\(
\begin{aligned}
& V_2=3 V_1=3 \times 2.24 \\
& =6.72 L
\end{aligned}
\)

Hint

(c) Work done under any thermodynamic process can be determined by area under the ‘p-V’ graph. As it can be observed maximum area is covered in option ‘c’.

Hint

(d)

\(
\Delta G=\Delta G^{\circ}+ RT \ln Q
\)
At equilibrium \(\Delta G=0, Q = K _{\text {eq }}\)
So, \(\Delta_r G^{\circ}=- RT\) in \(K _{\text {eq }}\)
\(
\Delta_r G^{\circ}=-8.314 J mol ^{-1} K^{-1} \times 300 K \times \ln \left(2 \times 10^{13}\right)
\)

Hint

(c) Given reaction,
\(
2 Cl ( g ) \rightarrow Cl _2(g)
\)
This is an endothermic reaction because it requires energy to break bonds.
So, the reverse reaction is exothermic
\(
\Delta_r H<0
\)
Also, two gaseous atoms combine together to form 1 gaseous molecule. So, the randomness,
\(
\Delta_r S<0
\)

Hint

(b)

\(
\begin{aligned}
&2 H ( g ) \rightarrow H _2(g)\\
&\text { Due to bond formation, entropy decreases. }
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& W =- P _{\text {ext }}\left( V _2- V _1\right) \\
& P _{\text {ext }}=2 \text { bar } \\
& V _1=0.1 L \\
& V _2=0.25 L \\
& W =-2 \text { bar }[0.25-0.1] L \\
& W =-2 \times 0.15 \text { bar } L \\
& W =-0.30 \text { bar } L \\
& W =(-0.30) \times 100=-30 J
\end{aligned}
\)

Hint

(a) Let bond dissociation energies of \(X_2, Y_2\) and \(X Y\) are
\(
x kJ mol ^{-1}, 0.5 \times kJ mol ^{-1} \text { and } xkJ mol ^{-1}
\)
\(
\frac{1}{2} X _2+\frac{1}{2} Y _2 \rightarrow XY ; \Delta H =-200 kJ mol ^{-1}
\)
\(
\Delta H=\Sigma( B . E )_{\text {Reactant }}-\Sigma( B . E )_{\text {Product }}
\)
\(
\begin{aligned}
&\Rightarrow-200=\left[\frac{1}{2} \times x+\frac{1}{2} \times 0.5 x\right]-[1 \times x]\\
&\text { On solving, } x=800 kJ mol ^{-1}
\end{aligned}
\)