Question

**Given:** $\text{C}_2\text{H}_4 + \text{H}_2 \rightleftharpoons \text{C}_2\text{H}_6$, $\Delta H = -32.7 \text{ kcal}$ (exothermic) To increase the equilibrium concentration of $\text{C}_2\text{H}_4$, the equilibrium must shift in the **reverse (left) direction** — except for option D, where direct addition raises concentration. Using **Le Chatelier's Principle**: **(A) Increasing temperature:** Since the forward reaction is exothermic, increasing temperature shifts equilibrium to the **left**, increasing $[\text{C}_2\text{H}_4]$. ✓ **(B) Decreasing pressure:** Moles of gas: Left side = 2 ($\text{C}_2\text{H}_4 + \text{H}_2$), Right side = 1 ($\text{C}_2\text{H}_6$). Decreasing pressure shifts equilibrium toward the side with **more moles of gas** (left), increasing $[\text{C}_2\text{H}_4]$. ✓ **(C) Removing some $\text{H}_2$:** Removing $\text{H}_2$ shifts equilibrium to the **left** to restore $[\text{H}_2]$, thereby increasing $[\text{C}_2\text{H}_4]$. ✓ **(D) Adding some $\text{C}_2\text{H}_4$:** Directly increases $[\text{C}_2\text{H}_4]$. Although equilibrium then shifts right consuming some $\text{C}_2\text{H}_4$, the final equilibrium concentration of $\text{C}_2\text{H}_4$ is still **higher** than the original. ✓ **Answer: (A), (B), (C), (D)**