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Standard free energy Questions in English
Class 11 Chemistry · 6-1.Equilibrium (Chemical Equilibrium) · Standard free energy
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101
DifficultMCQ
Calculate $\Delta G^\circ$ for the reaction, $CH_4(g) + H_2(g) \rightarrow C_2H_6(g)$ at $298 \text{ K}$, given $K_p = 2 \times 10^{17}$ and $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.
A
$-64.695 \text{ kJ mol}^{-1}$
B
$-98.716 \text{ kJ mol}^{-1}$
C
$-44.08 \text{ kJ mol}^{-1}$
D
$-58.78 \text{ kJ mol}^{-1}$
Solution
(B) The relationship between standard Gibbs free energy change and equilibrium constant is given by: $\Delta G^\circ = -RT \ln K_p$.
Given: $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$, $T = 298 \text{ K}$, $K_p = 2 \times 10^{17}$.
$\Delta G^\circ = -8.314 \times 298 \times \ln(2 \times 10^{17})$.
$\Delta G^\circ = -8.314 \times 298 \times (\ln 2 + \ln 10^{17})$.
$\Delta G^\circ = -8.314 \times 298 \times (0.693 + 17 \times 2.303)$.
$\Delta G^\circ = -8.314 \times 298 \times (0.693 + 39.151) = -8.314 \times 298 \times 39.844$.
$\Delta G^\circ \approx -98716 \text{ J mol}^{-1} = -98.716 \text{ kJ mol}^{-1}$.
Given: $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$, $T = 298 \text{ K}$, $K_p = 2 \times 10^{17}$.
$\Delta G^\circ = -8.314 \times 298 \times \ln(2 \times 10^{17})$.
$\Delta G^\circ = -8.314 \times 298 \times (\ln 2 + \ln 10^{17})$.
$\Delta G^\circ = -8.314 \times 298 \times (0.693 + 17 \times 2.303)$.
$\Delta G^\circ = -8.314 \times 298 \times (0.693 + 39.151) = -8.314 \times 298 \times 39.844$.
$\Delta G^\circ \approx -98716 \text{ J mol}^{-1} = -98.716 \text{ kJ mol}^{-1}$.
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View Solution102
DifficultMCQ
The equilibrium constant for a reaction is $100$. What will be the value of standard Gibbs energy change at $298 \text{ K}$? $(R = 8.314 \text{ J K}^{-1} \text{mol}^{-1})$
A
-$11.411$ \text{ kJ/mol}
B
-$5.744$ \text{ kJ/mol}
C
-$570.584$ \text{ kJ/mol}
D
-$57.058$ \text{ kJ/mol}
Solution
(A) The relationship between standard Gibbs energy change $(\Delta G^\circ)$ and equilibrium constant $(K)$ is given by: $\Delta G^\circ = -RT \ln K$.
Given: $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$, $T = 298 \text{ K}$, $K = 100$.
$\Delta G^\circ = -8.314 \times 298 \times \ln(100)$.
Since $\ln(100) = 2.303 \times \log_{10}(100) = 2.303 \times 2 = 4.606$.
$\Delta G^\circ = -8.314 \times 298 \times 4.606 \text{ J/mol}$.
$\Delta G^\circ = -11411.4 \text{ J/mol} = -11.411 \text{ kJ/mol}$.
Given: $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$, $T = 298 \text{ K}$, $K = 100$.
$\Delta G^\circ = -8.314 \times 298 \times \ln(100)$.
Since $\ln(100) = 2.303 \times \log_{10}(100) = 2.303 \times 2 = 4.606$.
$\Delta G^\circ = -8.314 \times 298 \times 4.606 \text{ J/mol}$.
$\Delta G^\circ = -11411.4 \text{ J/mol} = -11.411 \text{ kJ/mol}$.
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View Solution103
EasyMCQ
Identify the relation between the standard Gibbs free energy change $\Delta G^{\circ}$ and the equilibrium constant $K_c$ for a chemical reaction.
A
$\Delta G^{\circ} = RT \ln K_c$
B
$-\Delta G^{\circ} = \frac{RT}{\ln K_c}$
C
$\Delta G^{\circ} = \frac{\ln K_c}{RT}$
D
$-\Delta G^{\circ} = RT \ln K_c$
Solution
(D) The relationship between the standard Gibbs free energy change $\Delta G^{\circ}$ and the equilibrium constant $K_c$ is given by the equation:
$\Delta G^{\circ} = -RT \ln K_c$
Multiplying both sides by $-1$, we get:
$-\Delta G^{\circ} = RT \ln K_c$
Therefore, option $D$ is correct.
$\Delta G^{\circ} = -RT \ln K_c$
Multiplying both sides by $-1$, we get:
$-\Delta G^{\circ} = RT \ln K_c$
Therefore, option $D$ is correct.
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View Solution104
DifficultMCQ
Calculate the standard Gibbs energy change $(\Delta G^\circ)$ for a gaseous reaction at $298 \text{ K}$ if the equilibrium constant $K_p$ is $3.5 \times 10^{17}$ and the gas constant $R$ is $8.314 \text{ J K}^{-1} \text{mol}^{-1}$.
A
-$100.1 \text{ kJ mol}^{-1}$
B
-$79.5 \text{ kJ mol}^{-1}$
C
-$71.4 \text{ kJ mol}^{-1}$
D
-$89.5 \text{ kJ mol}^{-1}$
Solution
(A) The formula for standard Gibbs energy change is $\Delta G^\circ = -RT \ln K_p$.
Given: $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$, $T = 298 \text{ K}$, $K_p = 3.5 \times 10^{17}$.
Using the relation $\ln x = 2.303 \log_{10} x$:
$\Delta G^\circ = -2.303 \times R \times T \times \log_{10} K_p$.
$\Delta G^\circ = -2.303 \times 8.314 \times 298 \times \log_{10}(3.5 \times 10^{17})$.
$\log_{10}(3.5 \times 10^{17}) = \log_{10}(3.5) + 17 \approx 0.544 + 17 = 17.544$.
$\Delta G^\circ = -2.303 \times 8.314 \times 298 \times 17.544 \approx -100135 \text{ J mol}^{-1}$.
Converting to kJ: $\Delta G^\circ \approx -100.1 \text{ kJ mol}^{-1}$.
Given: $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$, $T = 298 \text{ K}$, $K_p = 3.5 \times 10^{17}$.
Using the relation $\ln x = 2.303 \log_{10} x$:
$\Delta G^\circ = -2.303 \times R \times T \times \log_{10} K_p$.
$\Delta G^\circ = -2.303 \times 8.314 \times 298 \times \log_{10}(3.5 \times 10^{17})$.
$\log_{10}(3.5 \times 10^{17}) = \log_{10}(3.5) + 17 \approx 0.544 + 17 = 17.544$.
$\Delta G^\circ = -2.303 \times 8.314 \times 298 \times 17.544 \approx -100135 \text{ J mol}^{-1}$.
Converting to kJ: $\Delta G^\circ \approx -100.1 \text{ kJ mol}^{-1}$.
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