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

Vedclass · Accepted Answer

$-887.00 \ kJ \ mol^{-1}$

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(A) The standard enthalpy change of reaction $\Delta_{r} H^{\circ}$ 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: $\Delta_{r} H^{\circ} = [\Delta_{f} H^{\circ}(CO_{2}) + 2 	imes \Delta_{f} H^{\circ}(H_{2}O)] - [\Delta_{f} H^{\circ}(CH_{4}) + 2 	imes \Delta_{f} H^{\circ}(O_{2})]$.Since $\Delta_{f} H^{\circ}(O_{2}) = 0 \ kJ \ mol^{-1}$ (standard state of an element),we substitute the given values:$\Delta_{r} H^{\circ} = [-390 + 2 	imes (-286)] - [-75 + 2 	imes 0]$$\Delta_{r} H^{\circ} = [-390 - 572] - [-75]$$\Delta_{r} H^{\circ} = -962 + 75 = -887 \ kJ \ mol^{-1}$.

Calculate the standard enthalpy change of the following reaction: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$ if $\Delta_{f} H^{\circ}(CH_{4}) = -75 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(CO_{2}) = -390 \ kJ \ mol^{-1}$,and $\Delta_{f} H^{\circ}(H_{2}O) = -286 \ kJ \ mol^{-1}$.

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