Mix Examples-Thermodynamics and Thermochemistry Questions in English

(B) To obtain the target reaction $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$,we manipulate the given equations:
$1$. Multiply equation $(A)$ by $4$: $4Na_{(s)} + 4H_2O_{(l)} \longrightarrow 4NaOH_{(s)} + 2H_{2(g)} \quad \Delta H_1 = 4 \times (-146) = -584 \ kJ$
$2$. Reverse equation $(B)$: $2NaOH_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)} + H_2O_{(l)} \quad \Delta H_2 = -418 \ kJ$
$3$. Keep equation $(C)$ as is: $2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H_3 = +259 \ kJ$
Adding these three equations:
$(4Na + 4H_2O) + (2NaOH + SO_3) + (2Na_2O + 2H_2) \longrightarrow (4NaOH + 2H_2) + (Na_2SO_4 + H_2O) + (4Na + 2H_2O)$
Simplifying the equation gives: $2Na_2O_{(s)} + 2SO_{3(g)} \longrightarrow 2Na_2SO_{4(s)}$
Dividing by $2$: $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$
The enthalpy change is $\Delta H = \frac{1}{2} (\Delta H_1 + \Delta H_2 + \Delta H_3) = \frac{1}{2} (-584 - 418 + 259) = \frac{1}{2} (-743) = -371.5 \ kJ$.
Wait,re-evaluating the combination:
$2 \times (A) + \frac{1}{2} \times (C) - (B) = 2(-146) + \frac{1}{2}(259) - 418 = -292 + 129.5 - 418 = -580.5 \ kJ \approx -581 \ kJ$.

(A) Given: $\Delta U = 2.1 \ kCal = 2100 \ cal$,$\Delta S = 20 \ cal \ K^{-1}$,$T = 25 + 273 = 298 \ K$,$R = 2 \ cal \ K^{-1} \ mol^{-1}$,$\Delta n_g = 2 - 0 = 2$.
$\Delta H = \Delta U + \Delta n_g RT = 2100 + (2 \times 2 \times 298) = 2100 + 1192 = 3292 \ cal$.
$\Delta G = \Delta H - T \Delta S = 3292 - (298 \times 20) = 3292 - 5960 = -2668 \ cal$.
$\Delta G = -2.668 \ kCal \approx -2.67 \ kCal$.

(A, B) Statement $(a)$ is correct: In $CrO_{5}$ (butterfly structure),the oxidation number of $Cr$ is $+6$.
Statement $(b)$ is correct: For the reaction $N_{2}O_{4(g)} \rightarrow 2NO_{2(g)}$,$\Delta n_{g} = 2 - 1 = 1$. Since $\Delta H = \Delta U + \Delta n_{g}RT$,and $\Delta n_{g} > 0$,we have $\Delta H > \Delta U$.
Statement $(c)$ is incorrect: For $0.1 \ N \ H_{2}SO_{4}$,$[H^{+}] = 0.1 \ M$,so $pH = -\log(0.1) = 1$. For $0.1 \ N \ HCl$,$[H^{+}] = 0.1 \ M$,so $pH = -\log(0.1) = 1$. Thus,$pH$ values are equal.
Statement $(d)$ is incorrect: At $25^{\circ} C$,$\frac{RT}{F} \approx 0.0257 \ V$. The value $0.0591 \ V$ corresponds to $\frac{2.303RT}{F}$.
Therefore,the correct statements are $(a)$ and $(b)$.

(B) The change in enthalpy $\Delta H$ for an ideal gas at constant pressure is given by the formula: $\Delta H = n C_{p} \Delta T$.
Given: $n = 2 \ mol$,$C_{p} = \frac{5}{2} R$,$T_{1} = 225^{\circ} C$,$T_{2} = 125^{\circ} C$.
Change in temperature $\Delta T = T_{2} - T_{1} = 125 - 225 = -100 \ K$.
Substituting the values: $\Delta H = 2 \times (\frac{5}{2} R) \times (-100) = 5 R \times (-100) = -500 R$.

(D) For an ideal gas,internal energy $(U)$ and enthalpy $(H)$ are functions of temperature only.
In an isothermal process,$\Delta T = 0$,which implies $\Delta U = 0$ and $\Delta H = 0$.
From the first law of thermodynamics,$\Delta U = Q + W$. Since $\Delta U = 0$,$Q = -W$.
For expansion,work is done,so $W \neq 0$,which means $Q \neq 0$.
Therefore,the correct combination is $\Delta U=0, Q \neq 0, W \neq 0$ and $\Delta H=0$.

(A) Step $1$: Calculate the millimoles of reactants.
$n(NaOH) = 20 \ mL \times 0.5 \ M = 10 \ mmol$.
$n(HCl) = 100 \ mL \times 0.1 \ M = 10 \ mmol$.
Step $2$: Determine the amount of water formed.
The reaction is $NaOH + HCl \rightarrow NaCl + H_2O$.
Since both reactants are $10 \ mmol$,$10 \ mmol$ of $H_2O$ is produced.
Step $3$: Calculate the heat of neutralization.
Heat released for $10 \ mmol$ $(0.01 \ mol)$ of $H_2O$ is $x \ kJ$.
Heat of neutralization is defined as the heat released per mole of $H_2O$ formed.
$\Delta H_{neut} = - \frac{x \ kJ}{0.01 \ mol} = -100 \ x \ kJ / mol$.

(B, C) For a process to be spontaneous,the total entropy change of the universe must be positive: $(\Delta S_{\text{system}}) + (\Delta S_{\text{surroundings}}) > 0$.
At constant temperature and pressure,the Gibbs free energy change must be negative: $(\Delta G_{\text{system}})_{T, p} < 0$.
At constant temperature and volume,the internal energy change must be negative: $(\Delta U_{\text{system}})_{T, V} < 0$.
Therefore,the correct statements are $B$ and $C$.

(C) For the reaction $X_2Y_{4(l)} \rightarrow 2XY_{2(g)}$
$\Delta n_g = \text{number of gaseous products} - \text{number of gaseous reactants} = 2 - 0 = 2$
Given,$\Delta U = 2 \ kcal = 2000 \ cal$,$\Delta S = 20 \ cal \ K^{-1}$,$T = 300 \ K$,and $R = 2 \ cal \ K^{-1} \ mol^{-1}$
Using the relation $\Delta H = \Delta U + \Delta n_g RT$
$\Delta H = 2000 + (2 \times 2 \times 300) = 2000 + 1200 = 3200 \ cal$
Now,using the Gibbs free energy equation $\Delta G = \Delta H - T \Delta S$
$\Delta G = 3200 - (300 \times 20) = 3200 - 6000 = -2800 \ cal$

(B) The spontaneity of a chemical reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T \Delta S$.
If $\Delta G < 0$,the reaction is spontaneous.
If $\Delta G > 0$,the reaction is non-spontaneous.
Therefore,the change in entropy $(\Delta S)$ along with the change in enthalpy $(\Delta H)$ determines the spontaneity of the reaction.

(A) For an isothermal process involving an ideal gas, the internal energy change $\Delta U$ and enthalpy change $\Delta H$ are zero because they depend only on temperature, which remains constant.
For an isothermal reversible expansion, the work done $w$ is given by $w = -nRT \ln(V_2/V_1) = -P_1V_1 \ln(P_1/P_2)$.
Given $P_1 = 0.5 \ MPa = 0.5 \times 10^6 \ Pa$, $V_1 = 20.0 \ dm^3 = 20.0 \times 10^{-3} \ m^3$, and $P_2 = 0.2 \ MPa$.
$w = -(0.5 \times 10^6 \ Pa) \times (20.0 \times 10^{-3} \ m^3) \times \ln(0.5/0.2)$.
$w = -10000 \ J \times \ln(2.5) = -10000 \times 2.303 \times \log(2.5)$.
Since $\log(2.5) = \log(5/2) = \log(5) - \log(2) = 0.6989 - 0.3010 = 0.3979$.
$w = -10000 \times 2.303 \times 0.3979 \approx -9166 \ J \approx -9.1 \ kJ$.
Since $\Delta U = q + w = 0$, then $q = -w = 9.1 \ kJ$.

(A) . Work done in reversible isothermal expansion: $W = -nRT \ln(\frac{V_2}{V_1}) = -2 \times 8.314 \times 300 \times \ln(10) \approx -11488 \ J = -11.5 \ kJ$. Magnitude is $11.5 \ kJ$ $(II)$.
$B$. Work done in irreversible isothermal expansion: $W = -P_{ext}(V_2 - V_1) = -3 \times 10^3 \ Pa \times (3 - 1) \ m^3 = -6000 \ J = -6 \ kJ$. Magnitude is $6 \ kJ$ $(III)$.
$C$. Change in internal energy: $\Delta U = nC_V \Delta T = 1 \times \frac{3}{2} \times 8.314 \times 320 \approx 3990 \ J \approx 4 \ kJ$. Magnitude is $4 \ kJ$ $(I)$.
$D$. Change in enthalpy: $\Delta H = nC_P \Delta T = 1 \times \frac{5}{2} \times 8.314 \times 337 \approx 7004 \ J \approx 7 \ kJ$. Magnitude is $7 \ kJ$ $(IV)$.
Thus,the correct match is $A-II, B-III, C-I, D-IV$.

(B) The standard enthalpy change of the reaction is calculated as:
$\Delta H^\circ = \Sigma \Delta_f H^\circ(\text{products}) - \Sigma \Delta_f H^\circ(\text{reactants})$
$\Delta H^\circ = 2 \times \Delta_f H^\circ(XY) - [\Delta_f H^\circ(X_2) + \Delta_f H^\circ(Y_2)]$
$\Delta H^\circ = 2(42) - (8 + 80) = 84 - 88 = -4 \ kJ \ mol^{-1} = -4000 \ J \ mol^{-1}$.
The standard entropy change of the reaction is:
$\Delta S^\circ = \Sigma S^\circ(\text{products}) - \Sigma S^\circ(\text{reactants})$
$\Delta S^\circ = 2 \times S^\circ(XY) - [S^\circ(X_2) + S^\circ(Y_2)]$
$\Delta S^\circ = 2(200) - (140 + 250) = 400 - 390 = 10 \ J \ K^{-1} \ mol^{-1}$.
Using the Gibbs-Helmholtz equation:
$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$
$\Delta G^\circ = -4000 \ J \ mol^{-1} - (600 \ K \times 10 \ J \ K^{-1} \ mol^{-1})$
$\Delta G^\circ = -4000 - 6000 = -10000 \ J \ mol^{-1} = -10 \ kJ \ mol^{-1}$.

(C) For an isothermal process,the magnitude of work done during compression is greater than the magnitude of work done during expansion between the same initial and final states.
In the case of expansion,the work done in multi-stage expansion $(w_B)$ is greater than in single-stage expansion $(w_A)$,so $|w_B| > |w_A|$.
In the case of compression,the work done in single-stage compression $(w_C)$ is greater than in multi-stage compression $(w_D)$,so $|w_C| > |w_D|$.
Comparing the two,the magnitude of work required for compression is always greater than the magnitude of work obtained during expansion between the same two states.
Therefore,the correct order of magnitude is $|w_C| > |w_D| > |w_B| > |w_A|$.

(D) Statement $I$ is false because for an ideal gas,the heat capacity at constant pressure $(C_p)$ is always greater than the heat capacity at constant volume $(C_v)$,as defined by the relation $C_p - C_v = R$.
Statement $II$ is true because in a constant volume process (isochoric process),the change in volume $\Delta V = 0$. Since work done $W = P\Delta V$,it follows that $W = 0$. According to the first law of thermodynamics,$\Delta U = Q + W$. Since $W = 0$,the heat added $(Q_v)$ is entirely used to increase the internal energy $(\Delta U)$,which corresponds to an increase in the chaotic motion of molecules and is reflected as a rise in temperature.

(C) . For a reversible isothermal expansion,the work done is $w = -nRT ln(V_2/V_1)$. Since $\Delta U = 0$ for an isothermal process,$q = -w = nRT ln(V_2/V_1)$. Thus,$A-II$.
$B$. In free expansion,the gas expands against zero external pressure $(P_{ext} = 0)$,so $w = 0$. For an isothermal process,$\Delta U = 0$,hence $q = 0$. Thus,$B-I$.
$C$. For an irreversible compression,the work done is $w = -P_{ext}(V_2 - V_1) = P_{ext}(V_1 - V_2)$. Thus,$C-III$.
$D$. For a cyclic reversible process,the entropy change of the system is zero $(\oint dS = 0)$. Since $dS = dq_{rev}/T$,it follows that $\oint \frac{dq_{rev}}{T} = 0$. Thus,$D-IV$.
The correct matching is $A-II, B-I, C-III, D-IV$.

(A) For the reaction $2A(g) + B(g) \rightarrow 2D(g)$,the change in the number of gaseous moles is $\Delta n_g = 2 - (2 + 1) = -1$.
We know the relation $\Delta H^{\circ} = \Delta U^{\circ} + \Delta n_g RT$.
Substituting the values: $\Delta H^{\circ} = -10 \times 10^3 \text{ J mol}^{-1} + (-1) \times 8.31 \text{ J K}^{-1} \text{ mol}^{-1} \times 298 \text{ K} = -10000 - 2476.38 = -12476.38 \text{ J mol}^{-1} = -12.476 \text{ kJ mol}^{-1}$.
Now,we use the Gibbs free energy equation: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.
$\Delta G^{\circ} = -12.476 \text{ kJ mol}^{-1} - 298 \text{ K} \times (-44 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}) = -12.476 + 13.112 = +0.636 \text{ kJ mol}^{-1}$.
Since $\Delta G^{\circ} > 0$,the reaction is non-spontaneous.