(C) Given: $\Delta H^o = -29.8 \, kJ \, mol^{-1}$,$\Delta S^o = -0.100 \, kJ \, K^{-1} \, mol^{-1}$,and $T = 298 \, K$. Using the Gibbs free energy eq

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(C) Given: $\Delta H^o = -29.8 \, kJ \, mol^{-1}$,$\Delta S^o = -0.100 \, kJ \, K^{-1} \, mol^{-1}$,and $T = 298 \, K$. Using the Gibbs free energy equation: $\Delta G^o = \Delta H^o - T\Delta S^o$. Substituting the values: $\Delta G^o = -29.8 \, kJ \, mol^{-1} - (298 \, K \times -0.100 \, kJ \, K^{-1} \, mol^{-1})$. $\Delta G^o = -29.8 + 29.8 = 0 \, kJ \, mol^{-1}$. Since $\Delta G^o = -RT \ln K_{eq}$,and $\Delta G^o = 0$,we have $0 = -RT \ln K_{eq}$. This implies $\ln K_{eq} = 0$,which means $K_{eq} = e^0 = 1$.

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

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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

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$1.0 \times 10^{-10}$
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$10$
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$1$
D
$1.0 \times 10^{10}$

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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

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

Given $\Delta G^o (HI, g) \cong + 1.7 \ kJ \ mol^{-1}$. What is the equilibrium constant at $25 ^oC$ for the reaction $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$?

For a system at equilibrium,$\Delta G = 0$ under which of the following conditions?

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

For a reaction,$\Delta G^{\circ} = -115 \, kJ$. What is the value of $\log \, K_p$ at $298 \, K$?

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