(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

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Standard free energy change is greater than zero $(\Delta G^o > 0)$

(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 $K_p 0)$.

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

Medium

For the equilibrium $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$ at $1 \ atm$ and $298 \ K$,which of the following statements is correct?

AIIMS 2004 All AIIMS

A
Standard free energy change is equal to zero $(\Delta G^o = 0)$
B
Free energy change is less than zero $(\Delta G < 0)$
C
Standard free energy change is less than zero $(\Delta G^o < 0)$
D
Standard free energy change is greater than zero $(\Delta G^o > 0)$

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6-1.Equilibrium (Chemical Equilibrium)

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Standard free energy

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AIIMS 2004 Paper All AIIMS years →

The equilibrium constant for the following reaction is $K_{p} = 3.44 \times 10^{24}$ at $25 ^{\circ}C$. Calculate the value of $\Delta_{f}G^{o}(SO_2)$. Given that the value of $\Delta_{f}G^{o}(SO_3)$ is $-88.52 \ kcal/mol$. The reaction is $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.

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If the change in standard Gibbs free energy for a reaction is less than $0$,then the value of the equilibrium constant $K_c$ is:

Easy

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For the reversible reaction,$A_{(s)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$,$\Delta G^{\circ} = -350 \ kJ$,which one of the following statements is true?

MediumKCET 2011

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At $298 \ K$, the equilibrium constant of the process $1.5 O_{2(g)} \rightleftharpoons O_{3(g)}$ is $3 \times 10^{-29}$. The standard free energy change (in $kJ \ mol^{-1}$) of the process is approximately ($R = 8.314 \ J \ mol^{-1} \ K^{-1}$; $\log 3 = 0.47$)

MediumTS EAMCET 2018

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For the reaction at $25\,^oC$,${N_2O_4}_{(g)} \rightleftharpoons 2NO_{2_{(g)}}$,if $\Delta G_f^o$ for $N_2O_4$ and $NO_2$ are $23.49 \, KCal$ and $12.39 \, KCal$ respectively,then $K_p$ for the reaction is: (in $, atm$)

Medium

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For the equilibrium $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$ at $1 \ atm$ and $298 \ K$,which of the following statements is correct?

Similar Questions

The equilibrium constant for the following reaction is $K_{p} = 3.44 \times 10^{24}$ at $25 ^{\circ}C$. Calculate the value of $\Delta_{f}G^{o}(SO_2)$. Given that the value of $\Delta_{f}G^{o}(SO_3)$ is $-88.52 \ kcal/mol$. The reaction is $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.

If the change in standard Gibbs free energy for a reaction is less than $0$,then the value of the equilibrium constant $K_c$ is:

For the reversible reaction,$A_{(s)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$,$\Delta G^{\circ} = -350 \ kJ$,which one of the following statements is true?

At $298 \ K$, the equilibrium constant of the process $1.5 O_{2(g)} \rightleftharpoons O_{3(g)}$ is $3 \times 10^{-29}$. The standard free energy change (in $kJ \ mol^{-1}$) of the process is approximately ($R = 8.314 \ J \ mol^{-1} \ K^{-1}$; $\log 3 = 0.47$)

For the reaction at $25\,^oC$,${N_2O_4}_{(g)} \rightleftharpoons 2NO_{2_{(g)}}$,if $\Delta G_f^o$ for $N_2O_4$ and $NO_2$ are $23.49 \, KCal$ and $12.39 \, KCal$ respectively,then $K_p$ for the reaction is: (in $, atm$)

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