Standard free energy Questions in English

(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^o = -RT \ln K$ or $\Delta G^o = -2.303 RT \log K$.
Given that $\Delta G^o = 0$,we substitute this into the equation: $0 = -2.303 RT \log K$.
Since $R$ (gas constant) and $T$ (temperature) are non-zero,it follows that $\log K = 0$.
Taking the antilog,we get $K = 10^0 = 1$.

(A) For a chemical reaction or a physical process,the criterion for equilibrium at constant temperature and pressure is that the change in Gibbs free energy is zero,i.e.,$\Delta G = 0$.

(B) At equilibrium,the Gibbs free energy change,$\Delta G$,is equal to $0$.
Therefore,a reaction attains equilibrium when the free energy change accompanying it is $0$.

(D) The reaction is $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$.
For the formation of $1 \ mol$ of $HI$,$\Delta G_f^o = + 1.7 \ kJ \ mol^{-1}$.
For the reaction $2HI \rightarrow H_2 + I_2$,the standard Gibbs free energy change is $\Delta G^o_{rxn} = [\Delta G_f^o(H_2) + \Delta G_f^o(I_2)] - 2 \times \Delta G_f^o(HI)$.
Since $H_2$ and $I_2$ are elements in their standard states,$\Delta G_f^o = 0$.
Therefore,$\Delta G^o_{rxn} = 0 - 2 \times (1.7 \ kJ \ mol^{-1}) = - 3.4 \ kJ \ mol^{-1} = - 3400 \ J \ mol^{-1}$.
Using the relation $\Delta G^o = - RT \ln K$,we have $\ln K = - \frac{\Delta G^o}{RT}$.
$\ln K = - \frac{- 3400}{8.314 \times 298} \approx \frac{3400}{2477.57} \approx 1.37$.
$K = e^{1.37} \approx 3.9$.
Note: If the question implies $\Delta G^o$ for the reaction is $+ 1.7 \ kJ$,then $\log K = - \frac{1700}{2.303 \times 8.314 \times 298} \approx - 0.298$,so $K = 10^{-0.298} \approx 0.5$.

(A) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^\circ = -RT \ln K_p = -2.303 \ RT \log K_p$.
Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,and $K_p = 2.47 \times 10^{-29}$.
Substituting the values: $\Delta G^\circ = -2.303 \times 8.314 \times 298 \times \log(2.47 \times 10^{-29})$.
$\Delta G^\circ = -2.303 \times 8.314 \times 298 \times (\log 2.47 + \log 10^{-29})$.
$\Delta G^\circ = -5705.85 \times (0.3927 - 29)$.
$\Delta G^\circ = -5705.85 \times (-28.6073) \approx 163215 \ J \ mol^{-1} \approx 163 \ kJ \ mol^{-1}$.

(D) For a reversible reaction at equilibrium,the change in Gibbs free energy $(\Delta G)$ is equal to $0$.
This is because at equilibrium,the rate of the forward reaction equals the rate of the backward reaction,and there is no net change in the system's state.

(A) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the formula: $\Delta G^o = -2.303 \ RT \ \log K$.
Given: $\Delta G^o = -4.606 \ kcal = -4606 \ cal$,$R = 2.0 \ cal \ mol^{-1} K^{-1}$,and $T = 227 + 273 = 500 \ K$.
Substituting the values: $-4606 = -2.303 \times 2.0 \times 500 \ \log K$.
$-4606 = -2303 \ \log K$.
$\log K = \frac{4606}{2303} = 2$.
Therefore,$K = 10^2 = 100$.

(D) The relationship between standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(K_c)$ is derived from the equation: $\Delta G = \Delta G^o + RT \ln Q$.
At equilibrium,the reaction quotient $Q = K_c$ and the change in Gibbs free energy $\Delta G = 0$.
Substituting these values into the equation gives: $0 = \Delta G^o + RT \ln K_c$.
Rearranging this,we get: $\Delta G^o = -RT \ln K_c$ or $-\Delta G^o = RT \ln K_c$.

(B) For the reaction $A + B \rightleftharpoons C + D$,the equilibrium constant $K_c$ is calculated as:
$K_c = \frac{[C][D]}{[A][B]} = \frac{10 \times 15}{3 \times 5} = \frac{150}{15} = 10$
At equilibrium,the Gibbs free energy change $\Delta G$ is given by the relation $\Delta G = -RT \ln K_c$ or $\Delta G = -2.303 \ RT \log_{10} K_c$.
Given $R = 2 \ cal \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,and $K_c = 10$:
$\Delta G = -2.303 \times 2 \times 300 \times \log_{10}(10)$
$\Delta G = -2.303 \times 600 \times 1 = -1381.8 \ cal$.

(B) The reaction is $\frac{1}{2} A_{2(g)} + \frac{3}{2} B_{2(g)} \rightarrow AB_{3(g)}$ with $\Delta H = -20 \, kJ = -20000 \, J$.
Calculate the change in entropy $(\Delta S)$:
$\Delta S = S_{AB_3} - (\frac{1}{2} S_{A_2} + \frac{3}{2} S_{B_2})$
$\Delta S = 50 - (\frac{1}{2} \times 60 + \frac{3}{2} \times 40)$
$\Delta S = 50 - (30 + 60) = 50 - 90 = -40 \, J K^{-1} mol^{-1}$.
At equilibrium,$\Delta G = 0$,so $\Delta H = T \Delta S$.
$T = \frac{\Delta H}{\Delta S} = \frac{-20000 \, J}{-40 \, J K^{-1}} = 500 \, K$.

(A) The relationship between the standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(K)$ is given by the equation: $\Delta G^o = -RT \ln K$.
Here,$R$ is the universal gas constant,$T$ is the absolute temperature in Kelvin,and $\ln K$ is the natural logarithm of the equilibrium constant.

(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^o = -RT \ln K_c$.
This can be rearranged as: $-\Delta G^o = RT \ln K_c$.
Therefore,the correct option is $B$.

(D) The relationship between standard free energy change and equilibrium constant is given by: $\Delta G^o = -RT \ln K$.
Since $\ln K = 2.303 \log_{10} K$,we can write: $\Delta G^o = -2.303 RT \log_{10} K$.
Rearranging for $\log_{10} K$: $\log_{10} K = \frac{-\Delta G^o}{2.303 RT}$.
Taking the antilog,we get: $K = 10^{\left( \frac{-\Delta G^o}{2.303 RT} \right)}$.

(D) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^o = -RT \ln K_p$.
Taking the exponential of both sides:
$\ln K_p = \frac{-\Delta G^o}{RT}$
$K_p = e^{\left( \frac{-\Delta G^o}{RT} \right)}$
Therefore,the correct option is $D$.

(B) The relationship between standard Gibbs free energy change and equilibrium constant is given by: $\Delta G^o = -2.303 \, RT \, \log \, K_{eq}$
Given: $R = 8 \, J \, K^{-1} \, mol^{-1}$,$T = 300 \, K$,$K_{eq} = 10$
Substituting the values: $\Delta G^o = -2.303 \times 8 \times 300 \times \log(10) \, J \, mol^{-1}$
Since $\log(10) = 1$: $\Delta G^o = -2.303 \times 8 \times 300 \, J \, mol^{-1} = -5527.2 \, J \, mol^{-1}$
Converting to $kJ \, mol^{-1}$: $\Delta G^o = -5.527 \, kJ \, mol^{-1}$

(D) The relationship between standard Gibbs energy change $\Delta G^{\circ}$ and equilibrium constant $K$ is given by $\Delta G^{\circ} = -RT \ln K$.
At equilibrium,$\Delta G = 0$,but $\Delta G^{\circ}$ is not necessarily zero unless $K = 1$.
Therefore,the Assertion is incorrect.
However,a chemical reaction is spontaneous in the direction of decreasing Gibbs energy $(\Delta G < 0)$ at constant temperature and pressure,which makes the Reason correct.

(B) The standard Gibbs free energy change is given by $\Delta_rG^o = \Delta_rH^o - T\Delta_rS^o$.
Substituting the values: $\Delta_rG^o = (-54.07 \times 10^3\, J\, mol^{-1}) - (298\, K \times 10\, J\, K^{-1}\, mol^{-1}) = -54070 - 2980 = -57050\, J\, mol^{-1}$.
We know that $\Delta_rG^o = -2.303 RT \log_{10}K$.
Substituting the values: $-57050 = -2.303 \times 8.314 \times 298 \times \log_{10}K$.
Given $2.303 \times 8.314 \times 298 = 5705$,we have $-57050 = -5705 \times \log_{10}K$.
Therefore,$\log_{10}K = \frac{57050}{5705} = 10$.

(B) The relationship between standard Gibbs free energy change $(\Delta G^\circ)$ and the equilibrium constant $(K_c)$ is given by the equation: $\Delta G^\circ = -2.303 \, RT \, \log K_c$.
If $\Delta G^\circ < 0$,then $-2.303 \, RT \, \log K_c < 0$.
This implies that $\log K_c > 0$,which means $K_c > 10^0 = 1$.
Therefore,the value of the equilibrium constant is greater than $1$.

(A) For a chemical reaction,the change in Gibbs free energy $(\Delta G)$ is related to the spontaneity of the process.
At equilibrium,the system is at its minimum Gibbs free energy state,meaning $\Delta G = 0$.
This condition specifically applies when the process is carried out at constant temperature $(T)$ and constant pressure $(P)$.

(A) The relationship between standard Gibbs free energy change $\Delta G^o$ and the equilibrium constant $K_p$ is given by the equation: $\Delta G^o = -RT \ln K_p$.
Given values are $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,and $K_p = 3 \times 10^{-29}$.
Substituting these values into the equation:
$\Delta G^o = -(8.314 \ J \ K^{-1} \ mol^{-1}) \times (298 \ K) \times \ln(3 \times 10^{-29})$.
$\Delta G^o = -2477.572 \times (\ln 3 + \ln 10^{-29})$.
$\Delta G^o = -2477.572 \times (1.0986 - 66.779)$.
$\Delta G^o = -2477.572 \times (-65.6804) \approx 162735 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$,we get $\Delta G^o \approx 162.74 \ kJ \ mol^{-1}$.

(B) The relationship between standard Gibbs free energy change $(\Delta G^{\circ})$ and the equilibrium constant $(K)$ is given by the equation: $\Delta G^{\circ} = -RT \ln K$.
Given that $\Delta G^{\circ} = 0$,we have $0 = -RT \ln K$.
Since $R$ (gas constant) and $T$ (temperature) are non-zero,$\ln K$ must be $0$.
Taking the exponential of both sides,$K = e^0 = 1$.

(A) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -2.303 \, RT \log \, K_p$.
Given: $\Delta G^{\circ} = -115 \, kJ = -115000 \, J$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$,and $T = 298 \, K$.
Substituting the values: $\log \, K_p = -\frac{\Delta G^{\circ}}{2.303 \, RT} = -\frac{-115000}{2.303 \times 8.314 \times 298}$.
$\log \, K_p = \frac{115000}{5705.84} \approx 20.155 \approx 20.16$.

(A) The relationship between the standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^o = -RT \ln K$ or $\Delta G^o = -2.303 \, RT \log K$.
Therefore,the equilibrium constant $K$ is directly related to the standard free energy change $\Delta G^o$.

(D) The relationship between the standard Gibbs free energy change $\Delta G^o$ and the equilibrium constant $K_p$ is given by the equation: $\Delta G^o = - RT \ln K_p$.
Rearranging this equation for $K_p$:
$\ln K_p = - \frac{\Delta G^o}{RT}$.
Taking the exponential of both sides:
$K_p = e^{- \frac{\Delta G^o}{RT}}$.
Therefore,the correct option is $D$.

(D) At $298 \ K$ and $1 \ atm$,liquid water is the stable phase,meaning the process $H_2O_{(l)} \rightarrow H_2O_{(g)}$ is non-spontaneous.
Since the process is non-spontaneous,the Gibbs free energy change $\Delta G$ for the forward reaction is greater than zero $(\Delta G > 0)$.
However,the question asks about the equilibrium state at these conditions.
Actually,at $298 \ K$ and $1 \ atm$,the equilibrium constant $K_p = P_{H_2O} = 0.0313 \ atm$.
Since $\Delta G^o = -RT \ln K_p$,and $K_p < 1$,$\Delta G^o$ must be positive $(\Delta G^o > 0)$.

(A) The relationship between Gibbs free energy change and the reaction quotient is given by $\Delta G = \Delta G^{\circ} + 2.303 \, RT \log Q$.
At equilibrium,the change in Gibbs free energy is zero,i.e.,$\Delta G = 0$,and the reaction quotient $Q$ becomes equal to the equilibrium constant $K$.
Substituting these values into the equation: $0 = \Delta G^{\circ} + 2.303 \, RT \log K$.
Rearranging the terms,we get $\Delta G^{\circ} = -2.303 \, RT \log K$.

(A) The process of vaporization is $X_{(l)} \rightleftharpoons X_{(g)}$.
For this process,the change in standard Gibbs free energy is $\Delta_rG^o = \Delta_fG^o [X_{(g)}] - \Delta_fG^o [X_{(l)}]$.
$\Delta_rG^o = -60.4 - (-65) = 4.6 \, kcal \, mol^{-1} = 4600 \, cal \, mol^{-1}$.
We know that $\Delta_rG^o = -RT \ln P$,where $P$ is the vapour pressure in $atm$.
$4600 = - (2 \, cal \, K^{-1} \, mol^{-1}) \times (500 \, K) \times \ln P$.
$4600 = -1000 \ln P$.
$\ln P = -4.6$.
Using $\ln P = 2.3 \log P$,we get $2.3 \log P = -4.6$.
$\log P = -2$.
$P = 10^{-2} \, atm = 0.01 \, atm$.

(A) The reaction is $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$.
At equilibrium,the total pressure $X$ is distributed according to the stoichiometric coefficients.
$P_{NH_3} = \frac{2}{3} X$ and $P_{CO_2} = \frac{1}{3} X$.
$K_p = (P_{NH_3})^2 (P_{CO_2}) = (\frac{2}{3} X)^2 (\frac{1}{3} X) = \frac{4}{9} X^2 \cdot \frac{1}{3} X = \frac{4}{27} X^3$.
Using the relation $\Delta_r G^o = -RT \ln K_p$:
$\Delta_r G^o = -RT \ln (\frac{4}{27} X^3)$.
Using logarithmic properties $\ln(ab) = \ln a + \ln b$ and $\ln(a^n) = n \ln a$:
$\Delta_r G^o = -RT (\ln X^3 + \ln \frac{4}{27}) = -RT (3 \ln X + \ln \frac{4}{27})$.

(A) The formula for standard Gibbs free energy change is $\Delta G^{\circ} = -2.303 RT \log K_{eq}$.
Given: $R = 8.314 \ J \cdot K^{-1} \cdot mol^{-1} = 8.314 \times 10^{-3} \ kJ \cdot K^{-1} \cdot mol^{-1}$,$T = 127 + 273 = 400 \ K$,and $K_{eq} = 10^5$.
Substituting the values:
$\Delta G^{\circ} = -2.303 \times (8.314 \times 10^{-3} \ kJ \cdot K^{-1} \cdot mol^{-1}) \times (400 \ K) \times \log(10^5)$
$\Delta G^{\circ} = -2.303 \times 8.314 \times 10^{-3} \times 400 \times 5$
$\Delta G^{\circ} = -2.303 \times 3.3256 \times 5$
$\Delta G^{\circ} = -38.294 \ kJ/mol$.

(B) For a reaction at equilibrium,the Gibbs free energy change is given by $\Delta G^o = \Delta H^o - T \Delta S^o = 0$.
Therefore,$T = \frac{\Delta H^o}{\Delta S^o}$.
Given $\Delta H^o = 25 \, kcal/mol = 25000 \, cal/mol$ and $\Delta S^o = 50 \, cal/K \cdot mol$.
Substituting the values: $T = \frac{25000 \, cal/mol}{50 \, cal/K \cdot mol} = 500 \, K$.

(D) The dissociation reaction is $A_2(g) \leftrightarrow 2A(g).$
Let the initial moles of $A_2$ be $1 \ mol.$
After $20 \%$ dissociation,moles of $A_2$ remaining $= 1 - 0.2 = 0.8 \ mol.$
Moles of $A$ formed $= 2 \times 0.2 = 0.4 \ mol.$
Total moles at equilibrium $= 0.8 + 0.4 = 1.2 \ mol.$
Partial pressures at $1 \ atm$ total pressure:
$P_{A_2} = \frac{0.8}{1.2} \times 1 = \frac{2}{3} \ atm.$
$P_A = \frac{0.4}{1.2} \times 1 = \frac{1}{3} \ atm.$
Equilibrium constant $K_p = \frac{(P_A)^2}{P_{A_2}} = \frac{(1/3)^2}{2/3} = \frac{1/9}{2/3} = \frac{1}{6}.$
Standard free energy change $\Delta G^o = -RT \ ln \ K_p = -8.314 \times 320 \times ln(1/6).$
$\Delta G^o = -8.314 \times 320 \times (-ln \ 6) = 8.314 \times 320 \times (ln \ 2 + ln \ 3).$
$\Delta G^o = 8.314 \times 320 \times (0.693 + 1.098) = 8.314 \times 320 \times 1.791 \approx 4763 \ J \ mol^{-1}.$

(C) Given: $\Delta H^o = -29.8 \, kJ \, mol^{-1}$,$\Delta S^o = -0.100 \, kJ \, K^{-1} \, mol^{-1}$,and $T = 298 \, K$.
Using the Gibbs free energy equation: $\Delta G^o = \Delta H^o - T\Delta S^o$.
Substituting the values: $\Delta G^o = -29.8 \, kJ \, mol^{-1} - (298 \, K \times -0.100 \, kJ \, K^{-1} \, mol^{-1})$.
$\Delta G^o = -29.8 + 29.8 = 0 \, kJ \, mol^{-1}$.
Since $\Delta G^o = -RT \ln K_{eq}$,and $\Delta G^o = 0$,we have $0 = -RT \ln K_{eq}$.
This implies $\ln K_{eq} = 0$,which means $K_{eq} = e^0 = 1$.

(D) The equilibrium condition is given by ${\Delta_r}{G^o} = 0$.
Setting the equation to zero: $120 - \frac{3}{8} \ T = 0$.
Solving for $T$: $T = \frac{120 \times 8}{3} = 320 \ K$.
For $T < 320 \ K$,${\Delta_r}{G^o} > 0$,which implies the reaction is non-spontaneous in the forward direction and $X$ is the major component.
For $T > 320 \ K$,${\Delta_r}{G^o} < 0$,which implies the reaction is spontaneous in the forward direction and $Y$ is the major component.
Checking the options:
$A$: $T = 300 \ K < 320 \ K$,so $X$ is major.
$B$: $T = 280 \ K < 320 \ K$,so $X$ is major.
$C$: $T = 350 \ K > 320 \ K$,so $Y$ is major.
$D$: $T = 315 \ K < 320 \ K$,so $X$ is major.
Thus,option $D$ is correct.

(B) The relationship between standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(K)$ is given by the equation: $\Delta G^o = -RT \ln K$ or $\Delta G^o = -2.303 \, RT \log K$.
$1$. If $\Delta G^o < 0$,then $\ln K > 0$,which implies $K > 1$ (Spontaneous reaction).
$2$. If $\Delta G^o = 0$,then $\ln K = 0$,which implies $K = 1$ (Equilibrium).
$3$. If $\Delta G^o > 0$,then $\ln K < 0$,which implies $K < 1$ (Non-spontaneous reaction).
Therefore,the match $\Delta G^o < 0, K < 1$ is $INCORRECT$.

(A) The standard Gibbs free energy change for the reaction is calculated as:
$\Delta G^o = \Delta G^o_f(NO_2) - [\Delta G^o_f(NO) + \frac{1}{2} \Delta G^o_f(O_2)]$
Substituting the given values:
$\Delta G^o = 52.0 - [87.0 + 0] = -35.0 \ kJ/mol = -35 \times 10^3 \ J/mol$
The relationship between standard Gibbs free energy and the equilibrium constant is:
$\Delta G^o = -RT \ln K_{eq}$
Rearranging for $\ln K_{eq}$:
$\ln K_{eq} = -\frac{\Delta G^o}{RT}$
Substituting the values:
$\ln K_{eq} = -\frac{-35 \times 10^3 \ J/mol}{8.314 \ J/mol \ K \times 298 \ K} = \frac{35 \times 10^3}{8.314 \times 298}$

(C) The standard Gibbs free energy change for the reaction is calculated as: $\Delta G^o = 2 \times \Delta G_f^o(NO_2) - \Delta G_f^o(N_2O_4)$
$\Delta G^o = 2 \times 12.39 - 23.49 = 1.29 \, KCal = 1290 \, Cal$
Using the relation $\Delta G^o = -RT \ln K_p$ or $\Delta G^o = -2.303 RT \log K_p$:
$1290 = -2.303 \times 1.987 \times 298 \times \log K_p$
$\log K_p = -1290 / (2.303 \times 1.987 \times 298) \approx -0.946$
$K_p = 10^{-0.946} \approx 0.1132 \, atm$

(NONE) The relationship between standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(Kp)$ is given by the equation: $\Delta G^o = -RT \ln Kp$.
Converting the natural logarithm $(\ln)$ to base $10$ logarithm $(\log)$,we get: $\Delta G^o = -2.303 \, RT \log Kp$.
Rearranging this for $\log Kp$,we get: $\log Kp = \frac{-\Delta G^o}{2.303 \, RT}$.
From $\Delta G^o = -RT \ln Kp$,we can write $\ln Kp = -\frac{\Delta G^o}{RT}$,which implies $Kp = e^{-\frac{\Delta G^o}{RT}}$.
Both options $A$ and $B$ represent the same correct mathematical expression $Kp = e^{-\frac{\Delta G^o}{RT}}$.
Since the question asks for the incorrect relation and all provided options $A, B, C,$ and $D$ are mathematically equivalent and correct representations of the thermodynamic relationship,there is no incorrect option provided.

(A) The relationship between standard Gibbs free energy change and equilibrium constant is given by: $\Delta G^{\circ} = -RT \ln K_p = -2.303 \, RT \log K_p$
Given: $R = 2.0 \, cal \, K^{-1} \, mol^{-1} = 2 \times 10^{-3} \, kcal \, K^{-1} \, mol^{-1}$,$T = 298 \, K$,$K_p = 10^{-8}$
Substituting the values: $\Delta G^{\circ} = -2.303 \times (2 \times 10^{-3} \, kcal \, K^{-1} \, mol^{-1}) \times 298 \, K \times \log(10^{-8})$
$\Delta G^{\circ} = -2.303 \times 2 \times 10^{-3} \times 298 \times (-8)$
$\Delta G^{\circ} = +10.97 \, kcal$ (approx $11.0 \, kcal$ or $10.05$ based on provided options calculation logic).

(A) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the formula: $\Delta_{r} G^{\ominus} = -RT \ln K_{p} = -2.303 \, RT \log K_{p}$.
Given values are $R = 8.314 \, J \, K^{-1} \, mol^{-1}$,$T = 298 \, K$,and $K_{p} = 2.47 \times 10^{-29}$.
Substituting these values into the equation:
$\Delta_{r} G^{\ominus} = -2.303 \times (8.314 \, J \, K^{-1} \, mol^{-1}) \times (298 \, K) \times \log(2.47 \times 10^{-29})$.
$\Delta_{r} G^{\ominus} = -2.303 \times 8.314 \times 298 \times (\log 2.47 + \log 10^{-29})$.
$\Delta_{r} G^{\ominus} = -5705.85 \times (0.3927 - 29)$.
$\Delta_{r} G^{\ominus} = -5705.85 \times (-28.6073) \approx 163229 \, J \, mol^{-1}$.
Converting to $kJ \, mol^{-1}$,we get $\Delta_{r} G^{\ominus} \approx 163 \, kJ \, mol^{-1}$.

(N/A) We know the relationship between the equilibrium constant $K$ and the standard Gibbs energy change $\Delta_{r} G^{\ominus}$ is given by:
$\Delta_{r} G^{\ominus} = -RT \ln K = -2.303 \, RT \log K$
Rearranging for $\log K$:
$\log K = \frac{-\Delta_{r} G^{\ominus}}{2.303 \, RT}$
Given:
$\Delta_{r} G^{\ominus} = -13.6 \, kJ \, mol^{-1} = -13.6 \times 10^{3} \, J \, mol^{-1}$
$R = 8.314 \, J \, K^{-1} \, mol^{-1}$
$T = 298 \, K$
Substituting the values:
$\log K = \frac{-(-13.6 \times 10^{3})}{2.303 \times 8.314 \times 298}$
$\log K = \frac{13600}{5705.84} \approx 2.3835$
$K = \text{antilog}(2.3835) \approx 2.42 \times 10^{2}$.

(N/A) The dissociation reaction is $N_{2}O_{4(g)} \rightleftharpoons 2NO_{2(g)}$.
If $N_{2}O_{4}$ is $50 \%$ dissociated,the mole fractions at equilibrium are:
$x_{N_{2}O_{4}} = \frac{1-0.5}{1+0.5} = \frac{0.5}{1.5} = \frac{1}{3}$
$x_{NO_{2}} = \frac{2 \times 0.5}{1+0.5} = \frac{1}{1.5} = \frac{2}{3}$
At $P = 1 \, atm$,the partial pressures are:
$p_{N_{2}O_{4}} = \frac{1}{3} \times 1 \, atm = 0.333 \, atm$
$p_{NO_{2}} = \frac{2}{3} \times 1 \, atm = 0.667 \, atm$
The equilibrium constant $K_{p}$ is:
$K_{p} = \frac{(p_{NO_{2}})^{2}}{p_{N_{2}O_{4}}} = \frac{(2/3)^{2}}{1/3} = \frac{4/9}{1/3} = \frac{4}{3} \approx 1.333 \, atm$
Using the relation $\Delta_{r} G^{\ominus} = -RT \ln K_{p}$ at $T = 333 \, K$:
$\Delta_{r} G^{\ominus} = -8.314 \, J K^{-1} mol^{-1} \times 333 \, K \times \ln(1.333)$
$\Delta_{r} G^{\ominus} = -8.314 \times 333 \times 0.2877 \approx -796.5 \, J mol^{-1}$

(A) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation:
$\Delta G^{\theta} = -RT \ln K_{eq}$
Alternatively,using base $10$ logarithm:
$\Delta G^{\theta} = -2.303 \, RT \log K_{eq}$
Given:
$R = 8.314 \, J \, K^{-1} \, mol^{-1}$
$T = 300 \, K$
$K_{eq} = 10$
Substituting the values:
$\Delta G^{\theta} = -(2.303) \times (8.314 \, J \, K^{-1} \, mol^{-1}) \times (300 \, K) \times \log(10)$
Since $\log(10) = 1$:
$\Delta G^{\theta} = -(2.303) \times (8.314) \times (300) \times 1 \, J \, mol^{-1}$
$\Delta G^{\theta} = -5744.14 \, J \, mol^{-1}$
Converting to $kJ \, mol^{-1}$:
$\Delta G^{\theta} = -5.744 \, kJ \, mol^{-1}$

(A) Given: $\Delta G^{\ominus} = 13.8 \, kJ \, mol^{-1} = 13.8 \times 10^{3} \, J \, mol^{-1}$,$T = 298 \, K$,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$.
The relationship between standard Gibbs free energy and equilibrium constant is given by: $\Delta G^{\ominus} = -RT \ln K_{c}$.
Rearranging for $\ln K_{c}$: $\ln K_{c} = -\frac{\Delta G^{\ominus}}{RT}$.
Substituting the values: $\ln K_{c} = -\frac{13.8 \times 10^{3}}{8.314 \times 298} = -\frac{13800}{2477.572} \approx -5.5698$.
Calculating $K_{c}$: $K_{c} = e^{-5.5698} \approx 3.81 \times 10^{-3}$.

(N/A) The standard Gibbs free energy change is given by the formula:
$\Delta G^{\ominus} = -RT \ln K_{c}$
Given:
$R = 8.314 \ J \ mol^{-1} \ K^{-1}$
$T = 300 \ K$
$K_{c} = 2 \times 10^{13}$
Substituting the values:
$\Delta G^{\ominus} = -8.314 \ J \ mol^{-1} \ K^{-1} \times 300 \ K \times \ln(2 \times 10^{13})$
$\Delta G^{\ominus} = -2494.2 \times (\ln 2 + \ln 10^{13})$
$\Delta G^{\ominus} = -2494.2 \times (0.693 + 13 \times 2.303)$
$\Delta G^{\ominus} = -2494.2 \times (0.693 + 29.939)$
$\Delta G^{\ominus} = -2494.2 \times 30.632$
$\Delta G^{\ominus} \approx -76400 \ J \ mol^{-1} = -76.4 \ kJ \ mol^{-1}$

(N/A) When reactions proceed in both directions,a dynamic equilibrium is established. This occurs only when the Gibbs free energy of the system is at a minimum.
The criterion for equilibrium for a reaction $A + B \rightleftharpoons C + D$ is $\Delta_{r} G = 0$.
Gibbs energy for a reaction where all reactants and products are in their standard states,$\Delta_{r} G^{\circ}$,is related to the equilibrium constant $K$ as follows:
$\Delta_{r} G = \Delta_{r} G^{\circ} + RT \ln Q$
At equilibrium,$\Delta_{r} G = 0$ and $Q = K$,therefore:
$0 = \Delta_{r} G^{\circ} + RT \ln K$
$\Delta_{r} G^{\circ} = -RT \ln K = -2.303 RT \log K$
Also,$\Delta_{r} G^{\circ} = \Delta_{r} H^{\circ} - T\Delta_{r} S^{\circ}$.
For strongly endothermic reactions,$\Delta_{r} H^{\circ} > 0$,which leads to $K < 1$,meaning the reaction does not favor product formation.
For exothermic reactions,$\Delta_{r} H^{\circ} < 0$,which leads to $K > 1$,meaning the reaction favors product formation.

(N/A) The relationship between standard Gibbs free energy change $\Delta_r G^o$ and equilibrium constant $K_p$ is given by the equation: $\Delta_r G^o = -RT \ln K_p$.
Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,$K_p = 6.022 \times 10^{-5}$.
Substituting the values: $\Delta_r G^o = -(8.314 \ J \ K^{-1} \ mol^{-1}) \times (298 \ K) \times \ln(6.022 \times 10^{-5})$.
$\Delta_r G^o = -2477.572 \times (\ln(6.022) + \ln(10^{-5}))$.
$\Delta_r G^o = -2477.572 \times (1.795 - 11.513)$.
$\Delta_r G^o = -2477.572 \times (-9.718) \approx 24077 \ J \ mol^{-1} = 24.08 \ kJ \ mol^{-1}$.

(A) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{o} = -RT \ln K$.
Given $\Delta G^{o} = -54.64 \ kcal = -54640 \ cal$,$R = 1.987 \ cal \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$.
Substituting the values: $-54640 = -(1.987) \times (298) \times \ln K$.
$\ln K = \frac{54640}{592.126} \approx 92.277$.
$K = e^{92.277} \approx 1.169 \times 10^{40}$.

The standard Gibbs free energy change for the reaction is given by $\Delta_{r}G^{o} = -RT \ln K_{p}$.
Given $R = 1.987 \times 10^{-3} \ kcal/(mol \cdot K)$,$T = 298 \ K$,and $K_{p} = 3.44 \times 10^{24}$.
$\Delta_{r}G^{o} = -(1.987 \times 10^{-3}) \times 298 \times \ln(3.44 \times 10^{24})$.
$\Delta_{r}G^{o} = -0.592 \times (\ln(3.44) + 24 \ln(10)) \approx -0.592 \times (1.235 + 55.288) \approx -34.5 \ kcal/mol$.
Also,$\Delta_{r}G^{o} = 2 \Delta_{f}G^{o}(SO_3) - 2 \Delta_{f}G^{o}(SO_2) - \Delta_{f}G^{o}(O_2)$.
Since $\Delta_{f}G^{o}(O_2) = 0$,we have $-34.5 = 2(-88.52) - 2 \Delta_{f}G^{o}(SO_2)$.
$-34.5 = -177.04 - 2 \Delta_{f}G^{o}(SO_2)$.
$2 \Delta_{f}G^{o}(SO_2) = -177.04 + 34.5 = -142.54$.
$\Delta_{f}G^{o}(SO_2) = -71.27 \ kcal/mol$.

(A) The standard Gibbs free energy change is given by $\Delta G^o = \Delta H^o - T\Delta S^o$.
Given $\Delta H^o = 18451.2 \ cal$ and $\Delta S^o = 29.16 \ cal/K$ at $T = 298 \ K$.
$\Delta G^o = 18451.2 - (298 \times 29.16) = 18451.2 - 8689.68 = 9761.52 \ cal$.
Using the relation $\Delta G^o = -RT \ln K_{eq}$,where $R = 1.987 \ cal/mol \cdot K$.
$\ln K_{eq} = -\frac{\Delta G^o}{RT} = -\frac{9761.52}{1.987 \times 298} = -\frac{9761.52}{592.126} \approx -16.485$.
$K_{eq} = e^{-16.485} \approx 6.94 \times 10^{-8}$.

(B) For the reaction,$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the equilibrium constant $K_p = 0.98$.
The relationship between standard Gibbs free energy change and the equilibrium constant is given by $\Delta_r G^{\circ} = -RT \ln K_p$ or $\Delta_r G^{\circ} = -2.303 \ RT \log K_p$.
Since $K_p = 0.98$,which is less than $1$,the value of $\log K_p$ is negative.
Therefore,$\Delta_r G^{\circ} = -2.303 \times R \times 298 \times (\text{negative value}) = \text{positive value}$.
$A$ positive value of $\Delta_r G^{\circ}$ indicates that the reaction is non-spontaneous in the forward direction under standard conditions.