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(A) The standard enthalpy change of the reaction is calculated using the formula: $\Delta_{r} H^{\circ} = \sum \Delta_{f} H^{\circ}(	ext{products}) -

Vedclass · Accepted Answer

$-891 \ kJ \ mol^{-1}$

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(A) The standard enthalpy change of the reaction is calculated using the formula: $\Delta_{r} H^{\circ} = \sum \Delta_{f} H^{\circ}(	ext{products}) - \sum \Delta_{f} H^{\circ}(	ext{reactants})$For the reaction $CH_{4(g)} + 2O_{2(g)} ightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$,the expression is:$\Delta_{r} H^{\circ} = [\Delta_{f} H^{\circ}(CO_2) + 2 	imes \Delta_{f} H^{\circ}(H_2O)] - [\Delta_{f} H^{\circ}(CH_4) + 2 	imes \Delta_{f} H^{\circ}(O_2)]$Since $O_2$ is an element in its standard state,$\Delta_{f} H^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$.Substituting the given values:$\Delta_{r} H^{\circ} = [-394 + 2 	imes (-286)] - [-75 + 0]$$\Delta_{r} H^{\circ} = [-394 - 572] - [-75]$$\Delta_{r} H^{\circ} = -966 + 75 = -891 \ kJ \ mol^{-1}$

Calculate the standard enthalpy change of the following reaction: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$ Given that: $\Delta_{f} H^{\circ}(CH_4) = -75 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(CO_2) = -394 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(H_2O) = -286 \ kJ \ mol^{-1}$

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