Calculate the enthalpy change when 12 g of c | vedclass

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(C) The reaction for the formation of methane $(CH_4)$ from its constituent elements is:$C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}$Given the standard

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$-75$

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(C) The reaction for the formation of methane $(CH_4)$ from its constituent elements is:$C_{(s)} + 2H_{2(g)} ightarrow CH_{4(g)}$Given the standard enthalpy of formation $(\Delta H_f^{\circ})$ of methane is $-75 \ kJ \ mol^{-1}$.This means the formation of $1 \ mol$ of $CH_4$ releases $75 \ kJ$ of energy.First,calculate the number of moles of carbon $(C)$ used:$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{12 \ g}{12 \ g \ mol^{-1}} = 1 \ mol$Since $1 \ mol$ of $C$ produces $1 \ mol$ of $CH_4$,the enthalpy change $(\Delta H)$ is:$\Delta H = n 	imes \Delta H_f^{\circ} = 1 \ mol 	imes (-75 \ kJ \ mol^{-1}) = -75 \ kJ$Therefore,the enthalpy change is $-75 \ kJ$.

Calculate the enthalpy change when $12 \ g$ of carbon reacts with sufficient hydrogen to form methane. If the enthalpy of formation of methane is $-75 \ kJ \ mol^{-1}$. (in $kJ$)

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