Question

Enthalpy of formation of methane is $-75\ \mathrm{kJ/mol}$. What is the enthalpy change for formation of 24 g of methane?

• A. $-112.5\ \mathrm{kJ}$
• B. $-75\ \mathrm{kJ}$
• C. $-150\ \mathrm{kJ}$
• D. $-130\ \mathrm{kJ}$

Hint:

Since $24\ \mathrm{g}$ is exactly $1.5$ times the molar mass of methane ($16\ \mathrm{g}$), the heat change must simply be $1.5$ times the baseline molar value. Calculation shortcut: $-75 + (-37.5) = -112.5\ \mathrm{kJ}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the standard molar enthalpy of formation for methane ($\Delta H_f^{\circ} = -75\ \mathrm{kJ/mol}$), which is the heat released when 1 mole of methane is produced. We need to compute the net enthalpy change when $24\ \mathrm{g}$ of methane is formed.

Step 2: Key Formula or Approach:
First, find the number of moles ($n$) of the substance using its given mass ($m$) and its molar mass ($M$): $$n = \frac{m}{M}$$ Then, scale the total enthalpy change ($\Delta H$) by multiplying the number of moles by the molar enthalpy of formation: $$\Delta H = n \times \Delta H_f^{\circ}$$

Step 3: Detailed Explanation:
Determine the molar mass of methane ($\mathrm{CH}_4$): $$M = 12\ (\text{for Carbon}) + 4 \times 1\ (\text{for Hydrogen}) = 16\ \mathrm{g/mol}$$ Calculate the number of moles contained in $24\ \mathrm{g}$ of methane: $$n = \frac{24\ \mathrm{g}}{16\ \mathrm{g/mol}} = 1.5\ \mathrm{mol}$$ Now calculate the total enthalpy change for the production of $1.5\ \mathrm{mol}$: $$\Delta H = 1.5\ \mathrm{mol} \times (-75\ \mathrm{kJ/mol})$$ $$\Delta H = -112.5\ \mathrm{kJ}$$

Step 4: Final Answer:
The total enthalpy change is $-112.5\ \mathrm{kJ}$, which corresponds directly with option (A).