Chemical EquilibriumHardBloom L3

Question

For the reaction $$\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$$ at a given temperature, the equilibrium amount of $\text{H}_2(g)$ can be increased by:

Options

A.Adding a suitable catalyst
B.Adding an inert gas
C.Decreasing the volume of the container
D.Increasing the amount of $\text{CO}(g)$

Solution

**Given:** $\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$ **Key Observation:** The number of moles of gas on both sides is equal ($\Delta n_g = 0$). **Analysis of each option using Le Chatelier's Principle:** **(A) Adding a catalyst:** A catalyst only increases the rate of reaching equilibrium; it does not shift the equilibrium position. ❌ **(B) Adding an inert gas** (at constant volume): Does not change the partial pressures of reactants or products. No shift in equilibrium. ❌ **(C) Decreasing the volume:** Increases total pressure. Since $\Delta n_g = 0$, a pressure change has no effect on the equilibrium position. ❌ **(D) Increasing the amount of $\text{CO}(g)$:** Increases the concentration of a reactant. By Le Chatelier's Principle, the equilibrium shifts in the **forward direction**, producing more $\text{CO}_2(g)$ and $\text{H}_2(g)$. ✓ **Answer: (D)**

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