(C) The standard free energy change is given by the formula: $\Delta G^{\circ} = -RT \ln K = -2.303 RT \log K$. Given: $R = 8.314 \ J \ mol^{-1} \ K^{

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(C) The standard free energy change is given by the formula: $\Delta G^{\circ} = -RT \ln K = -2.303 RT \log K$. Given: $R = 8.314 \ J \ mol^{-1} \ K^{-1}$,$T = 25 + 273 = 298 \ K$,and $K = 3.8 \times 10^{-3}$. Substituting the values: $\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times \log (3.8 \times 10^{-3})$. Since $\log (3.8 \times 10^{-3}) = -2.42$,we have: $\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times (-2.42)$. $\Delta G^{\circ} \approx 13817 \ J \ mol^{-1} \approx 13.8 \ kJ \ mol^{-1}$.

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

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If the equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{\circ} C$,what is the standard free energy change of the process? $(R = 8.314 \ J \ mol^{-1} \ K^{-1}, \log 0.0038 = -2.42)$

TS EAMCET 2018 All TS EAMCET

A
$5.7 \ kJ \ mol^{-1}$
B
$9.9 \ kJ \ mol^{-1}$
C
$13.8 \ kJ \ mol^{-1}$
D
$15.6 \ kJ \ mol^{-1}$

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6-1.Equilibrium (Chemical Equilibrium)

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$2O_{3(g)} \rightleftharpoons 3O_{2(g)}$
At $300 \ K$,ozone is $50\%$ dissociated. The standard free energy change at this temperature and $1 \ atm$ pressure is $(-) \dots \ J \ mol^{-1}$ (Nearest integer).
[Given: $\ln 1.35 = 0.3$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ ]

MediumJEE MAIN 2022

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Find out $\ln K_{eq}$ for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \ K$.
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
Given:
$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$
$\Delta G^o_f (NO) = 87.0 \ kJ/mol$
$\Delta G^o_f (O_2) = 0 \ kJ/mol$

Difficult

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Consider the following reaction at $298 \ K$.
$\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)} ; K_{P} = 2.47 \times 10^{-29}$.
$\Delta_{r} G^{\ominus}$ for the reaction is $ . . . . . . \ kJ$. (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

DifficultJEE MAIN 2024

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Write the formula relating the equilibrium constant $K$ and the standard Gibbs free energy change $\Delta G^{\circ}$.

Easy

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

Easy

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If the equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{\circ} C$,what is the standard free energy change of the process? $(R = 8.314 \ J \ mol^{-1} \ K^{-1}, \log 0.0038 = -2.42)$

Similar Questions

$2O_{3(g)} \rightleftharpoons 3O_{2(g)}$At $300 \ K$,ozone is $50\%$ dissociated. The standard free energy change at this temperature and $1 \ atm$ pressure is $(-) \dots \ J \ mol^{-1}$ (Nearest integer).[Given: $\ln 1.35 = 0.3$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ ]

Find out $\ln K_{eq}$ for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \ K$.$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$Given:$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$$\Delta G^o_f (NO) = 87.0 \ kJ/mol$$\Delta G^o_f (O_2) = 0 \ kJ/mol$

Consider the following reaction at $298 \ K$.$\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)} ; K_{P} = 2.47 \times 10^{-29}$.$\Delta_{r} G^{\ominus}$ for the reaction is $ . . . . . . \ kJ$. (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

Write the formula relating the equilibrium constant $K$ and the standard Gibbs free energy change $\Delta G^{\circ}$.

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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$2O_{3(g)} \rightleftharpoons 3O_{2(g)}$
At $300 \ K$,ozone is $50\%$ dissociated. The standard free energy change at this temperature and $1 \ atm$ pressure is $(-) \dots \ J \ mol^{-1}$ (Nearest integer).
[Given: $\ln 1.35 = 0.3$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ ]

Find out $\ln K_{eq}$ for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \ K$.
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
Given:
$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$
$\Delta G^o_f (NO) = 87.0 \ kJ/mol$
$\Delta G^o_f (O_2) = 0 \ kJ/mol$

Consider the following reaction at $298 \ K$.
$\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)} ; K_{P} = 2.47 \times 10^{-29}$.
$\Delta_{r} G^{\ominus}$ for the reaction is $ . . . . . . \ kJ$. (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

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