Q51 / 180 · COMEDK 2024 (Afternoon Shift) · Chemistry
$\text{S}_8$ on heating at a temperature above 1000 K, changes to $\text{S}_2(g)$. When $\text{S}_8$ is heated above 1000 K at an initial pressure of 1 atm, the partial pressure of $\text{S}_8$ falls by 32% at equilibrium. The equilibrium constant for the conversion is:
- A$2.55 \mathrm{~atm}^3$
- Correct answerB$3.94 \mathrm{~atm}^3$
- C$3.20 \mathrm{~atm}^3$
- D$4.50 \mathrm{~atm}^3$
Solution
$S_8(g) \rightleftharpoons 4S_2(g)$
Initial pressure of $S_8 = 1$ atm. Pressure falls by 32%:
$P_{S_8} = 1 - 0.32 = 0.68$ atm
$P_{S_2} = 4 \times 0.32 = 1.28$ atm
$K_p = \dfrac{(P_{S_2})^4}{P_{S_8}} = \dfrac{(1.28)^4}{0.68} = \dfrac{2.684}{0.68} \approx 3.94$ atm$^3$
Initial pressure of $S_8 = 1$ atm. Pressure falls by 32%:
$P_{S_8} = 1 - 0.32 = 0.68$ atm
$P_{S_2} = 4 \times 0.32 = 1.28$ atm
$K_p = \dfrac{(P_{S_2})^4}{P_{S_8}} = \dfrac{(1.28)^4}{0.68} = \dfrac{2.684}{0.68} \approx 3.94$ atm$^3$
Hence, the correct answer is option: 'B'.