Calculate the enthalpy change when 12 g carbon react with sufficient hydrogen to form methane. If enthalpy…
Calculate the enthalpy change when 12 g carbon react with sufficient hydrogen to form methane. If enthalpy of formation of methane is $-75 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
$\quad-45 \mathrm{~kJ}$
-60 kJ
-75 kJ
-90 kJ
Solution
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 $\mathrm{C}:-75 \mathrm{~kJ}$.