(C) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta_{r}G^{\circ} = -RT \ln K

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(C) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta_{r}G^{\circ} = -RT \ln K_p$ Given that $T = 300 \ K$,$K_p = 100.0$,and $\ln 10 = 2.3$. Substituting the values: $\Delta_{r}G^{\circ} = -R \times 300 \times \ln(100)$ Since $\ln(100) = \ln(10^2) = 2 \ln(10) = 2 \times 2.3 = 4.6$. Therefore,$\Delta_{r}G^{\circ} = -R \times 300 \times 4.6 = -1380 R$. Comparing this with $-xR$,we get $x = 1380$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Standard free energy

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For the reaction $A_{(g)} \rightarrow B_{(g)},$ the value of the equilibrium constant at $300 \ K$ and $1 \ atm$ is equal to $100.0.$ The value of $\Delta_{r}G^{\circ}$ for the reaction at $300 \ K$ and $1 \ atm$ in $J \ mol^{-1}$ is $-xR,$ where $x$ is ........... (Rounded off to the nearest integer) ($R = 8.31 \ J \ mol^{-1} K^{-1}$ and $\ln 10 = 2.3$)

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A
$1255$
B
$1460$
C
$1380$
D
$1290$

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6-1.Equilibrium (Chemical Equilibrium)

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The equilibrium concentrations of the species in the reaction $A + B \rightleftharpoons C + D$ are $2, 3, 10$ and $6 \, mol \, L^{-1}$,respectively at $300 \, K$. $\Delta G^{\circ}$ for the reaction is $(R = 2 \, cal \, mol^{-1} \, K^{-1})$ (in $, cal$)

DifficultNEET 2023

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For the reaction,$A_{(g)} + B_{(g)} \to C_{(g)} + D_{(g)}$,$\Delta H^o$ and $\Delta S^o$ are,respectively,$-29.8 \, kJ \, mol^{-1}$ and $-0.100 \, kJ \, K^{-1} \, mol^{-1}$ at $298 \, K$. The equilibrium constant for the reaction at $298 \, K$ is

DifficultJEE MAIN 2016

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In the equilibrium state,the value of $\Delta G$ is:

EasyKCET 2007

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If $\Delta_fG^o [X_{(l)}] = -65 \, kcal \, mol^{-1}$ and $\Delta_fG^o [X_{(g)}] = -60.4 \, kcal \, mol^{-1},$ the vapour pressure of $X$ at $500 \, K$ would be about ...... $atm$.
Given: $R = 2 \, cal \, K^{-1} \, mol^{-1}$,$\ln \, a = 2.3 \, \log \, a$.

Medium

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Find the equilibrium constant for the reaction below at $25$ $^{\circ}C$:
$H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_{2}O_{(g)}$ [ $\Delta_{f}G^{o} = -54.64 \ kcal$ ].

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The equilibrium concentrations of the species in the reaction $A + B \rightleftharpoons C + D$ are $2, 3, 10$ and $6 \, mol \, L^{-1}$,respectively at $300 \, K$. $\Delta G^{\circ}$ for the reaction is $(R = 2 \, cal \, mol^{-1} \, K^{-1})$ (in $, cal$)

For the reaction,$A_{(g)} + B_{(g)} \to C_{(g)} + D_{(g)}$,$\Delta H^o$ and $\Delta S^o$ are,respectively,$-29.8 \, kJ \, mol^{-1}$ and $-0.100 \, kJ \, K^{-1} \, mol^{-1}$ at $298 \, K$. The equilibrium constant for the reaction at $298 \, K$ is

In the equilibrium state,the value of $\Delta G$ is:

If $\Delta_fG^o [X_{(l)}] = -65 \, kcal \, mol^{-1}$ and $\Delta_fG^o [X_{(g)}] = -60.4 \, kcal \, mol^{-1},$ the vapour pressure of $X$ at $500 \, K$ would be about ...... $atm$.Given: $R = 2 \, cal \, K^{-1} \, mol^{-1}$,$\ln \, a = 2.3 \, \log \, a$.

Find the equilibrium constant for the reaction below at $25$ $^{\circ}C$:$H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_{2}O_{(g)}$ [ $\Delta_{f}G^{o} = -54.64 \ kcal$ ].

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If $\Delta_fG^o [X_{(l)}] = -65 \, kcal \, mol^{-1}$ and $\Delta_fG^o [X_{(g)}] = -60.4 \, kcal \, mol^{-1},$ the vapour pressure of $X$ at $500 \, K$ would be about ...... $atm$.
Given: $R = 2 \, cal \, K^{-1} \, mol^{-1}$,$\ln \, a = 2.3 \, \log \, a$.

Find the equilibrium constant for the reaction below at $25$ $^{\circ}C$:
$H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_{2}O_{(g)}$ [ $\Delta_{f}G^{o} = -54.64 \ kcal$ ].