Q122 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Chemistry
What is the work done in the formation of 1 mole gaseous methane from solid carbon and gaseous hydrogen at $27^\circ\text{C}$? ($R=8.314\ \text{J k}^{-1}\text{ mol}^{-1}$)
- A$-2.4942$ kJ
- Correct answerB$+2.4942$ kJ
- C$-24.942$ kJ
- D$+24.942$ kJ
Solution
The balanced chemical equation for the formation of $1$ mole of gaseous methane is:
$\text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}$
The change in the number of moles of gaseous substances ($\Delta n_g$) is:
$\Delta n_g = n_{g,\text{products}} - n_{g,\text{reactants}}$
$\Delta n_g = 1 - 2 = -1$
The work done ($W$) during a chemical reaction at constant temperature and pressure is given by:
$W = -\Delta n_g R T$
Given:
$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$
$T = 27^\circ\text{C} = 27 + 273 = 300 \text{ K}$
Substituting the values:
$W = -(-1) \times 8.314 \times 300$
$W = +2494.2 \text{ J}$
$W = +2.4942 \text{ kJ}$
Answer: $+2.4942$ kJ
$\text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}$
The change in the number of moles of gaseous substances ($\Delta n_g$) is:
$\Delta n_g = n_{g,\text{products}} - n_{g,\text{reactants}}$
$\Delta n_g = 1 - 2 = -1$
The work done ($W$) during a chemical reaction at constant temperature and pressure is given by:
$W = -\Delta n_g R T$
Given:
$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$
$T = 27^\circ\text{C} = 27 + 273 = 300 \text{ K}$
Substituting the values:
$W = -(-1) \times 8.314 \times 300$
$W = +2494.2 \text{ J}$
$W = +2.4942 \text{ kJ}$
Answer: $+2.4942$ kJ
Hence, the correct answer is option: 'B'.