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

The reaction under consideration is the formation of methane \(\left(\mathrm{CH}_4\right)\) from carbon \((\mathrm{C})\) and hydrogen \(\left(\mathrm{H}_2\right)\), which can be represented as:
\(\mathrm{C}(\mathrm{~s})+2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_4(\mathrm{~g})\)

The given standard enthalpy of formation for methane \(\left(\mathrm{CH}_4\right)\) is:
\(\Delta H_f^{\circ}=-75 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

This value indicates the enthalpy change for the formation of one mole of methane under standard conditions.
To calculate the enthalpy change for the reaction involving 12 g of carbon, follow these steps:
Calculate the number of moles of carbon (C):
The molar mass of carbon is approximately \(12 \mathrm{~g} / \mathrm{mol}\). Therefore, for 12 g of carbon, the number of moles is:
\(\text { moles of } \mathrm{C}=\frac{\text { mass }}{\text { molar mass }}=\frac{12 \mathrm{~g}}{12 \mathrm{~g} / \mathrm{mol}}=1 \mathrm{~mol}\)

Use the number of moles of carbon to find the enthalpy change:
Since 1 mole of carbon produces 1 mole of methane and the enthalpy of formation of methane is \(-75 \mathrm{~kJ} \mathrm{~mol}^{-1}\), the total enthalpy change for the reaction is:
\(\Delta H=-75 \mathrm{~kJ} / \mathrm{mol} \times 1 \mathrm{~mol}=-75 \mathrm{~kJ}\)

Thus, the enthalpy change when 12 g of carbon react with sufficient hydrogen to form methane is -75 kJ .
Correct answer is Option \(\mathrm{C}:-75 \mathrm{~kJ}\).