Thermodynamics

Heat at constant pressure,

$ \begin{aligned} & q_p=n C_p \Delta T \\ & \Rightarrow \quad 1 \times 10.314 \times 1=10.314 \mathrm{~J} \end{aligned} $

Enthalpy change $\Delta H$

$ \Delta H=q_p=10.314 \mathrm{~J} / \mathrm{mol} $

Given, $\Delta S=-42.4 \mathrm{JK}^{-1}$

$ \Delta H=-41.2 \mathrm{~kJ}=-41200 \mathrm{~J} $

Using the equation, $\Delta G=\Delta H-T \Delta S$

$ T_{\mathrm{eq}}=\frac{-41200 \mathrm{~J}}{-42.4 \mathrm{JK}^{-1}} \approx 971.8 \mathrm{~K} $

A reversible process is one that can be reversed by an infinitesimal change in a condition, allowing the system and surroundings to be restored to their original states. Such process is occur at equilibrium.

I. Vaporisation of a liquid at its boiling point This is a reversible process because the liquid and vapour are in equilibrium at the boiling point.

II. Expansion of gas into vacuum This is an irreversible, spontaneous process as the system is not in equilibrium and cannot be reversed without external work.

III. Transformation of a solid into liquid at its melting point This is a reversible processes as the solid and liquid phases are in equilibrium at the melting point.

IV. Neutralisation of an acid by a base This is an irreversible, spontaneous reaction that proceeds to completion.

Therefore, only processes I and III are reversible.

$ \text { Gibbs' free energy is given by } $

$ \begin{array}{rlrl} \Delta G^{\circ} & =-R T \ln (K) \\ \Rightarrow \quad-115000 \mathrm{~J} & =-8.314 \times 298 \times \ln K_p \\ \Rightarrow \quad \log _{10}\left(K_p\right) & =\frac{46.42}{2.303} \\ & \log _{10}\left(K_p\right) =+20.15 \end{array} $

$ \begin{aligned} &\text { Work done in isothermal reversible expansion, }\\ &\begin{aligned} W & =-n R T \ln \left(\frac{p_2}{p_1}\right) \\ W & =-(1)(8.3)(300) \ln \left(\frac{2}{20}\right)=5730 \mathrm{~J} / \mathrm{mole} \text { or } 5.73 \mathrm{~kJ} / \mathrm{mol} \\ x & =5.73 \end{aligned} \end{aligned} $

$ \text { } \begin{array} GGiven,\,\,{r} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^{\circ}=-286 \mathrm{~kJ}\,\,\,\,\,\,\,\,\,\,\, \ldots(\mathrm{i}) \end{array} $

$ \begin{aligned} \mathrm{C}(s)+\mathrm{O}_2(g) & \longrightarrow \mathrm{CO}_2(g);\Delta H^{\circ} =-394 \mathrm{~kJ} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$ \mathrm{CH}_4(g)+2 \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^{\circ}=-890 \mathrm{~kJ}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

$ \begin{aligned} &\text { Reverse reaction given in Eq. (ii), }\\ &\begin{aligned} & \mathrm{CO}_2(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{~s})+\mathrm{O}_2(g) ; \Delta H^{\circ}=+394 \mathrm{~kJ} \,\,\,\,\,\ldots(\mathrm{iv}) \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Now, add Eq. (iii) and (iv), }\\ &\begin{gathered} \mathrm{CH}_4(g)+2 \mathrm{O}_2(g)+\mathrm{CO}_2(g) \longrightarrow \mathrm{CO}_2(s)+2 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{C}(s)+\mathrm{O}_2(g) \\ \mathrm{CH}_4(g)+\mathrm{O}_2(g) \longrightarrow \mathrm{C}(s)+2 \mathrm{H}_2 \mathrm{O}(l) \\ \text { So, } \Delta H^{\circ}=-890+394=-496 \mathrm{~kJ} \end{gathered} \end{aligned} $

Statement-I

Both internal energy (U) and work (W) are state functions.

• Internal energy (U) is indeed a state function — it depends only on the state (e.g., temperature, pressure, volume) of the system, not on the path taken.

• Work (W), on the other hand, is not a state function; it depends on the path taken during a process. Different paths between the same initial and final states can result in different amounts of work.

✅ So, Statement-I is incorrect.

Statement-II

During the free expansion of an ideal gas into vacuum, the work done is zero.

In free expansion:

• The gas expands into a vacuum.

• There is no external pressure acting on the gas.

• Therefore, $ W = \int P_{\text{ext}} dV = 0 $.

✅ Hence, Statement-II is correct.

✅ Correct answer: Option D

Statement-I is not correct, but Statement-II is correct.

For a reaction to be spontaneous at all temperature, the enthalpy change ( $\Delta H^{\circ}$ ) must be negative (exothermic) and the entropy change ( $\Delta S^{\circ}$ ) must be positive (increase in disorderness).

Given graph

$C \rightarrow A$ : Temperature is constant, volume increases on increasing pressure.

Thus, isothermal.

$B \rightarrow C$ : Isobaric, pressure is constant, by increasing temperature, volume decreases.

$A \rightarrow B$ : By increasing temperature, volume remains same, thus isochoric.

In isothermal change (process), internal energy of ideal gas, $\Delta U=0$

$ \begin{aligned} p V & =n R T \\ \Delta(p V) & =0, \Delta H=\Delta U+\Delta(p V) \\ \Delta H & =0 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

Extensive property : mass, volume and enthalpy.

Intensive property : density, temperature and pressure.

Thus there are 3 extensive property and intensive property.

The complete combustion of ethanol is as follows

$ \begin{aligned} & \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)+3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l) \\ & \Delta_r H^{\ominus}=\Sigma n \Delta_f H_{\text {product }}^{\ominus}-\Sigma m \Delta_f H_{\text {reactant }}^{\ominus} \\ & =[(2 \times-393)+(3 \times-286)] -[(1 \times-277)+(3 \times 0)] \end{aligned} $

$ =-1367 \mathrm{~kJ} $

Statement given in option (b) is incorrect. Its correct form is,

The spontaneity of a reaction with a negative enthalpy change (exothermic) and a negative entropy change (decreasing disorder) depends on temperature.

When $\Delta H$ is negative and $\Delta S$ is negative the reaction is non-spontaneous at high temperature.

First, we need to find out how many moles of water we have.

$ \begin{aligned} n & = \frac{m}{M} = \frac{180}{18} \\ \\ & = 10~\mathrm{mol} \\ \Delta T & = T_2 - T_1 = 15 - 10 = 5~\mathrm{K} \end{aligned} $

The energy needed to heat the water is given by the formula: $Q = n C_p \Delta T$

Here, $n$ is number of moles (10), $C_p$ is the specific heat ($75.3~\mathrm{J~mol}^{-1}~\mathrm{K}^{-1}$), and $\Delta T$ is the temperature change (5 K).

So, $Q = 10 \times 75.3 \times 5 = 3765~\mathrm{J}$

The statement given in option II and III are incorrect.

While statement I is correct.

The correct of form statement II and III is,

• Enthalpy is an extensive property.

• For the process, $\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{H}_2 \mathrm{O}(s)$ the entropy decreases.

The complete combustion reaction is

$ \begin{aligned} & \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(s)+6 \mathrm{O}_2(g) \rightarrow 6 \mathrm{CO}_2(g)+6 \mathrm{H}_2 0(n) \\ & \Delta H_{\text {comb }}^{\circ}=\Sigma \Delta H_{\text {froduct }}^{\circ}-\Sigma \Delta H_{\text {frectant }}^{\circ} \\ & =[6 \times(-393)+6 \times(-286)]-[1 \times(-1170) +6 \times 0]\end{aligned} $

$ \begin{aligned} &=-2904 \mathrm{~kJ}\\ &\begin{aligned} \text { Moles of glucose } & =\frac{\text { Mass }}{\text { Molar mass }}=\frac{18 \mathrm{~g}}{180} \\ & =0.1 \mathrm{~mol} \end{aligned}\\ &\begin{aligned} \text { Heat liberated } & =\text { Moles } \times \Delta H_{\text {combustion }}^{\circ} \\ & =0.1 \times(-2904) \\ & \approx-290 \mathrm{~kJ} \end{aligned} \end{aligned} $

The reaction in which $n_{\text {product }} \neq n_{\text {reactant }}$ will have $\Delta H \neq \Delta U$ So, among the given reactions, in reaction,

$ \begin{aligned} & \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \longrightarrow 2 \mathrm{NH}_3(g) \\ & \Delta U \neq \Delta H \\ & \text { As, } n_{\text {product }}=2, n_{\text {reactant }}=4 \\ & \Delta n_g=2-4=-2 \\ & \Delta H=\Delta U+\Delta n_g R T \\ & \Delta H=\Delta U-2 R T \end{aligned} $

Given $\Delta_r U^{\circ}=-10.5 \mathrm{~kJ}$

$ \begin{aligned} & \Delta_r S^{\circ}=+44.1 \mathrm{~J} / \mathrm{K} \\ & \Delta_r G^{\circ}=? \end{aligned} $

Using formula

$ \begin{aligned} & \Delta H^{\circ}=\Delta U^{\circ}+\Delta(p V) \\ & \Delta H^{\circ}=\Delta U^{\circ}+p \Delta V+V \Delta p \end{aligned} $

as $\Delta p=0$

So, $\Delta H^{\circ}=\Delta U^{\circ}+p \Delta V$

Now, as $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$

$ \Delta G^{\circ}=\Delta U^{\circ}+p \Delta V-T \Delta S^{\circ} $

$ \begin{aligned} & \quad \begin{aligned} p \Delta V & =-R T \\ \because \quad p \Delta V & =\Delta n_g R T \\ \Delta n_g & =(2-3)=-1 \\ \text { So, } \Delta G^{\circ} & =\Delta U^{\circ}-R T-T \Delta S^{\circ} \\ = & -10.5-\left(8.314 \times 10^{-3} \times 298\right)-(298 \left. \times 44.1 \times 10^{-3}\right) \end{aligned} \\ & \quad \approx-26.119 \mathrm{~kJ} \end{aligned} $

The Gibbs free energy is given by

$ \begin{aligned} & \Delta G=\Delta H-T \Delta S \\ & \Delta S=\frac{\Delta H-\Delta G}{T} \end{aligned} $

Substituting the values,

$ \begin{aligned} & =\frac{-24000 \mathrm{~J}-(-9000 \mathrm{~J})}{298 \mathrm{~K}} \\ \Delta S & =-50.33 \mathrm{~J} / \mathrm{K} \end{aligned} $

The balanced equation for the combustion of ethanol is,

$ \begin{aligned} & \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O} \\ & \Delta H_{\text {reaction }}=\Sigma \Delta H_{f \text { product }}-\Sigma \Delta H_{\text {reactant }} \\ & =\left[2 \Delta H_{f \mathrm{CO}_2}+3 \Delta H_{f \mathrm{H}_2 \mathrm{O}}\right]-\left[\Delta H_{f \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}\right. \left.+3 \Delta H_{f \mathrm{O}_2}\right] \\ & \quad=[2 y+3 z]-[x+0] \\ & \quad=2 y+3 z-x \end{aligned} $

Statement I: Work is a path function.

• True.

Work depends on the path taken between initial and final states, not just on the states themselves. Hence, work is a path function.

Statement II: Enthalpy is an extensive property.

• True.

Enthalpy ($H$) depends on the amount of substance. If the system size doubles, enthalpy doubles. So it is an extensive property.

Statement III: Lattice enthalpy of ionic compounds can be obtained from Born–Haber cycle.

• True.

The Born–Haber cycle uses Hess’s law to calculate lattice enthalpy indirectly from measurable thermodynamic quantities.

✅ Correct statements: II and III are correct, and actually, I is also correct.

Wait, let’s double-check statement I:

Work is NOT a state function (true), but it is a path function — the statement says "Work is a path function" — That is correct.

Therefore all three I, II, and III are correct.

✅ Final Answer: Option D — I, II and III

Let's analyze each process in terms of entropy change ($\Delta S$):

I. Sublimation of dry ice (solid CO₂ → gas CO₂)

• A solid goes directly to a gas.

• The disorder (randomness) increases significantly.

• Entropy increases → $\Delta S > 0$

II. Freezing of water (liquid → solid)

• A liquid becomes a more ordered solid structure.

• Disorder decreases.

• Entropy decreases → $\Delta S < 0$

III. Crystallisation of the dissolved substance

• Dissolved ions/molecules (disordered) form a regular crystal lattice (ordered).

• Disorder decreases.

• Entropy decreases → $\Delta S < 0$

IV. Burning of rocket fuel (combustion)

• Chemical reaction produces gases and increases randomness overall.

• Entropy increases → $\Delta S > 0$

✅ Processes with negative entropy change:

II. Freezing of water

III. Crystallisation of the dissolved substance

✅ Correct answer: Option B — II and III only

Statement I

During isothermal expansion of an ideal gas its enthalpy decreases.

For an ideal gas, enthalpy $ H = U + pV $.

For an ideal gas:

$ U = n C_V T $

and since internal energy $U$ depends only on temperature for an ideal gas,

$ dU = 0 \quad \text{if temperature is constant (isothermal process)}. $

Also, $ pV = nRT$.

Therefore,

$ H = U + pV = nC_VT + nRT = n(C_V + R)T = nC_PT $

Since $T$ is constant (isothermal), enthalpy $H$ is also constant.

Hence, enthalpy does not decrease, it remains the same.

✅ Statement I is incorrect.

Statement II

When 2.0 L of an ideal gas expands isothermally into vacuum, $ \Delta U = 0 $.

For free expansion into vacuum:

• External pressure $ = 0$

• So $ w = 0 $

• No heat exchange ($ q = 0 $) because the system is isolated (expanding into vacuum)

• Thus, $ \Delta U = q + w = 0 + 0 = 0$

Also, for an ideal gas, $ \Delta U $ depends only on temperature, and since it is isothermal, $ \Delta U = 0 $.

✅ Statement II is correct.

✅ Final Answer

Statement I is not correct but Statement II is correct.

Correct option: D

Given:

$ m = 180\ \mathrm{g} $

$ \Delta T = 15 - 10 = 5\ \mathrm{K} $

$ q = 3765\ \mathrm{J} $

$ M(\mathrm{H_2O}) = 18\ \mathrm{g/mol} $

We need to find $ C_p $ in J mol⁻¹ K⁻¹.

Step 1: Use the relation $ q = n C_p \Delta T $

$ n = \frac{m}{M} = \frac{180}{18} = 10\ \mathrm{mol} $

Substitute values:

$ 3765 = 10 \times C_p \times 5 $

Step 2: Solve for $ C_p $

$ C_p = \frac{3765}{10 \times 5} = \frac{3765}{50} = 75.3\ \mathrm{J\ mol^{-1}\ K^{-1}} $

✅ Final Answer:

$ \boxed{C_p = 75.3\ \mathrm{J\ mol^{-1}\ K^{-1}}} $

Correct Option: A (75.3)

Number of moles of hydrogen

$ \begin{aligned} & =\frac{\text { mass }}{\text { molar mass }} \\ & =\frac{10 \mathrm{~g}}{2 \mathrm{~g} / \mathrm{mol}}=5 \mathrm{~mol} \end{aligned} $

Using work done in isothermal condition,

$ \begin{aligned} & W=-n R T \ln \left[\frac{p_1}{p_2}\right] \\ & W=-5 \times 8.3 \times 273 \times \ln \left[\frac{10}{1}\right] \\ & W \simeq-26090 \mathrm{~J}=-26.09 \mathrm{~kJ} \end{aligned} $

In a cyclic process, net heat absorbed = work done $=$ area enclosed in $p V$ diagram.

Area $=(3-1) \mathrm{atm} \times(4-2) \mathrm{L}$

$=2 \mathrm{~atm} \times 2 \mathrm{~L}=4 \mathrm{~L} \mathrm{~atm}$.

Now, convert to energy units if needed,

$4 \mathrm{Latm} \times 101.3 \mathrm{~J}=\sim 405 \mathrm{~J}$ (optional)

But in $\mathrm{L} \cdot \mathrm{atm}$, answer is 4 L atm

Let, $A_2=x$, then $B_2=4 x, A B=2 x$

$ \begin{aligned} \Delta H & =\left[\frac{1}{2} A_2+\frac{1}{2} B_2 \longrightarrow A B\right] \\ & =\left(\frac{1}{2} x+2 x\right)-2 x=\frac{1}{2} x=-100 \rightarrow x=-200 \end{aligned} $

So, $B_2=4 x=-800 \mathrm{~kJ} / \mathrm{mol}$

Enthalpy of neutralisation of strong acids and bases is $-57 \mathrm{~kJ} / \mathrm{mol}$. which is fixed for reaction of 1 mole of $\mathrm{H}^{+}$ with 1 mole of $\mathrm{OH}^{-}$ to form 1 mole of water.

Let's analyze what happens during a free (or unrestrained) expansion of an ideal gas under adiabatic conditions:

• Adiabatic process : By definition, in an adiabatic process, no heat is transferred between the system and its surroundings. Thus, $ q = 0 $.


• Free expansion : In a free expansion, the gas expands into a vacuum or against zero external pressure. Since work ($ w $) is defined as the product of pressure ($ P $) and volume change ($ \Delta V $), and since the external pressure is zero, we conclude that $ w = 0 $.


• Temperature change ($ \Delta T $) : Since work is zero and no heat is transferred, the first law of thermodynamics tells us that the change in internal energy ($ \Delta U $) is zero as well. For an ideal gas, internal energy is a function of temperature only; thus, if the internal energy does not change, the temperature does not change. Therefore, $ \Delta T = 0 $.

Based on the above points, the free expansion of an ideal gas under adiabatic conditions is characterized by no heat transfer ($ q = 0 $), no work done ($ w = 0 $), and no change in temperature ($ \Delta T = 0 $). Therefore, the correct option is:

Option A : $ q = 0, \Delta T = 0, w = 0 $

The statement that is not correct among the given options is Option A.

Option A states that $\Delta \mathrm{G}$ is positive for a spontaneous reaction, which is incorrect. The criterion for spontaneity in a chemical reaction is based on the Gibbs free energy change ($\Delta \mathrm{G}$) for the process. A spontaneous reaction is one that occurs without needing continuous input of energy from an external source. The correct relation is that for a spontaneous reaction, $\Delta \mathrm{G}$ is negative ($\Delta \mathrm{G} < 0$). When $\Delta \mathrm{G}$ is negative, it indicates that the process can occur spontaneously under the given conditions.

Option B accurately states that $\Delta \mathrm{G}$ is positive for a non-spontaneous reaction. A positive value of $\Delta \mathrm{G}$ ($\Delta \mathrm{G} > 0$) indicates that the reaction is not spontaneous under the given conditions and requires external energy to proceed.

Option C is correct in stating that $\Delta \mathrm{G}$ is zero for a reversible reaction at equilibrium. When a reaction is at equilibrium, it has reached a state where the forward and reverse reactions occur at equal rates, and there is no net change in the composition of the system. At this point, the Gibbs free energy is at its minimum for the given conditions, and $\Delta \mathrm{G} = 0$, indicating that the system is in a state of maximum stability and no further net change can occur without the input or removal of energy.

Option D correctly states that $\Delta \mathrm{G}$ is negative for a spontaneous reaction. As explained previously, a negative Gibbs free energy change signifies that a process or reaction can proceed spontaneously in the direction written.