The Delta U^o of combustion of CH_{4(g)} at a | vedclass

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(B) The balanced chemical equation for the combustion of methane is:$CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$Calculate the change in

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$< \Delta U^o$

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(B) The balanced chemical equation for the combustion of methane is:$CH_{4(g)} + 2O_{2(g)} ightarrow CO_{2(g)} + 2H_2O_{(l)}$Calculate the change in the number of gaseous moles,$\Delta n_g$:$\Delta n_g = (n_{products, gas}) - (n_{reactants, gas}) = 1 - (1 + 2) = 1 - 3 = -2$Using the relation between enthalpy change and internal energy change:$\Delta H^o = \Delta U^o + \Delta n_g RT$Substituting the value of $\Delta n_g$:$\Delta H^o = \Delta U^o + (-2)RT$$\Delta H^o = \Delta U^o - 2RT$Since $2RT$ is a positive value,subtracting it from $\Delta U^o$ results in a value smaller than $\Delta U^o$:$	herefore \Delta H^o < \Delta U^o$

The $\Delta U^o$ of combustion of $CH_{4(g)}$ at a certain temperature is $-393 \ kJ/mol$. Then the value of $\Delta H^o$ is:

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