Q30 / 150 · MHT CET 2026 (19 April Shift 2) · Chemistry
Calculate the work done for the following reaction at $27\,^\circ\text{C}$
$\text{C}_2\text{H}_{4(g)} + \text{H}_{2(g)} \longrightarrow \text{C}_2\text{H}_{6(g)}$ ($R = 8.314\,\text{JK}^{-1}\text{mol}^{-1}$)
$\text{C}_2\text{H}_{4(g)} + \text{H}_{2(g)} \longrightarrow \text{C}_2\text{H}_{6(g)}$ ($R = 8.314\,\text{JK}^{-1}\text{mol}^{-1}$)
- Correct answerA$2494.2$ J
- B$124.71$ J
- C$3741.3$ J
- D$187.07$ J
Solution
The given reaction is $\text{C}_2\text{H}_{4(g)} + \text{H}_{2(g)} \longrightarrow \text{C}_2\text{H}_{6(g)}$
Change in number of moles of gaseous species, $\Delta n_g = n_p - n_r$
$\Delta n_g = 1 - (1 + 1) = -1$
Work done, $W = -\Delta n_g RT$
$W = -(-1) \times 8.314 \times (27 + 273)$
$W = 1 \times 8.314 \times 300$
$W = 2494.2$ J
Answer: $2494.2$ J
Change in number of moles of gaseous species, $\Delta n_g = n_p - n_r$
$\Delta n_g = 1 - (1 + 1) = -1$
Work done, $W = -\Delta n_g RT$
$W = -(-1) \times 8.314 \times (27 + 273)$
$W = 1 \times 8.314 \times 300$
$W = 2494.2$ J
Answer: $2494.2$ J
Hence, the correct answer is option: 'A'.