For the reaction CH_{4(g)} + 2O_{2(g)} rightl | vedclass

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(C) $1$. The reaction is exothermic because $\Delta H$ is negative $(-170.8 \ kJ \ mol^{-1})$.$2$. The expression for the equilibrium constant $K_p$ i

Vedclass · Accepted Answer

The reaction is spontaneous at all temperatures.

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(C) $1$. The reaction is exothermic because $\Delta H$ is negative $(-170.8 \ kJ \ mol^{-1})$.$2$. The expression for the equilibrium constant $K_p$ is given by the ratio of the partial pressures of products to reactants raised to their stoichiometric coefficients: $K_p = \frac{p_{CO_2} 	imes (p_{H_2O})^2}{p_{CH_4} 	imes (p_{O_2})^2}$.$3$. The reaction involves $3 \ mol$ of gaseous reactants and $3 \ mol$ of gaseous products,so $\Delta n_g = 3 - 3 = 0$. The entropy change $\Delta S$ is approximately zero,so the spontaneity depends on the temperature $(T)$ according to $\Delta G = \Delta H - T\Delta S$. Thus,it is not spontaneous at all temperatures.

For the reaction $CH_{4(g)} + 2O_{2(g)} \rightleftharpoons CO_{2(g)} + 2H_{2}O_{(g)}$ with $\Delta H = -170.8 \ kJ \ mol^{-1}$,which of the following statements is incorrect?

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