Find out ln K_{eq} for the formation of NO_2 | vedclass

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(A) The standard Gibbs free energy change for the reaction is calculated as:$\Delta G^o = \Delta G^o_f(NO_2) - [\Delta G^o_f(NO) + \frac{1}{2} \Delta

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$\frac{35 	imes 10^3}{8.314 	imes 298}$

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(A) The standard Gibbs free energy change for the reaction is calculated as:$\Delta G^o = \Delta G^o_f(NO_2) - [\Delta G^o_f(NO) + \frac{1}{2} \Delta G^o_f(O_2)]$Substituting the given values:$\Delta G^o = 52.0 - [87.0 + 0] = -35.0 \ kJ/mol = -35 	imes 10^3 \ J/mol$The relationship between standard Gibbs free energy and the equilibrium constant is:$\Delta G^o = -RT \ln K_{eq}$Rearranging for $\ln K_{eq}$:$\ln K_{eq} = -\frac{\Delta G^o}{RT}$Substituting the values:$\ln K_{eq} = -\frac{-35 	imes 10^3 \ J/mol}{8.314 \ J/mol \ K 	imes 298 \ K} = \frac{35 	imes 10^3}{8.314 	imes 298}$

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:
$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$
$\Delta G^o_f (NO) = 87.0 \ kJ/mol$
$\Delta G^o_f (O_2) = 0 \ kJ/mol$

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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$