(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 standard Gibbs free energy change $\Delta G^{\circ}$ at $25^{\circ} C$ for the dissociation of $N_2O_{4(g)}$ to $NO_{2(g)}$ is (given, equilibrium constant $K_{eq} = 0.15, R = 8.314 \ J \ K^{-1} \ mol^{-1}$) (in $kJ$)

MediumWBJEE 2013

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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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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 $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.

Easy

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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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The $INCORRECT$ match in the following is

MediumJEE MAIN 2019

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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 standard Gibbs free energy change $\Delta G^{\circ}$ at $25^{\circ} C$ for the dissociation of $N_2O_{4(g)}$ to $NO_{2(g)}$ is (given, equilibrium constant $K_{eq} = 0.15, R = 8.314 \ J \ K^{-1} \ mol^{-1}$) (in $kJ$)

$2O_{3(g)} \rightleftharpoons 3O_{2(g)}$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 $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ ]

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 $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.

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 $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

The $INCORRECT$ match in the following is

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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 $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ ]

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 $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)