For the reaction 2 H_{2(g)} + O_{2(g)} longri | vedclass

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(D) The relationship between Gibbs free energy,enthalpy,and entropy is given by the equation: $\Delta G = \Delta H - T\Delta S$. Rearranging for entro

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$-44 \ J \ K^{-1}$

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(D) The relationship between Gibbs free energy,enthalpy,and entropy is given by the equation: $\Delta G = \Delta H - T\Delta S$. Rearranging for entropy change: $\Delta S = \frac{\Delta H - \Delta G}{T}$. Given: $\Delta H = -241.60 \ kJ \ mol^{-1} = -241600 \ J \ mol^{-1}$,$\Delta G = -228.4 \ kJ \ mol^{-1} = -228400 \ J \ mol^{-1}$,and $T = 300 \ K$. Substituting the values: $\Delta S = \frac{-241600 - (-228400)}{300} \ J \ K^{-1}$. $\Delta S = \frac{-241600 + 228400}{300} \ J \ K^{-1} = \frac{-13200}{300} \ J \ K^{-1} = -44 \ J \ K^{-1}$.

For the reaction $2 H_{2(g)} + O_{2(g)} \longrightarrow 2 H_2O_{(g)}$,at $300 \ K$,$\Delta G$ and $\Delta H$ of water are $-228.4 \ kJ \ mol^{-1}$ and $-241.60 \ kJ \ mol^{-1}$,respectively. Calculate the value of change in entropy for the given reaction.

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