Reaction 1: SO2(g) + 1/2 O2(g) <-> SO3(g) K1 Reaction 2: 2 SO3(g) <-> 2 SO2(g) + O2(g) K2 Which of t — Equilibrium Chemistry Question

STEPS:

1. Write the equilibrium constant expression ($K_c$) for Reaction 1 based on the law of mass action:
The balanced equation for Reaction 1 is:
$$\text{SO}_2(g) + \frac{1}{2}\ \text{O}_2(g) \rightleftharpoons \text{SO}_3(g) \quad$$
The equilibrium constant expression ($K_1$) for this reaction is:
$$K_1 = \frac{[\text{SO}_3]}{[\text{SO}_2][\text{O}_2]^{1/2}}$$

2. Write the equilibrium constant expression ($K_c$) for Reaction 2:
The balanced equation for Reaction 2 is:
$$2\ \text{SO}_3(g) \rightleftharpoons 2\ \text{SO}_2(g) + \text{O}_2(g) \quad$$
The equilibrium constant expression ($K_2$) for this reaction is:
$$K_2 = \frac{[\text{SO}_2]^2[\text{O}_2]}{[\text{SO}_3]^2}$$

3. Analyze the relationship by manipulating the chemical equations:
To transform Reaction 1 into Reaction 2, two successive mathematical manipulations must be performed:
* Step A: Reverse the reaction. Reversing Reaction 1 swaps the positions of the reactants and products:
$$\text{SO}_3(g) \rightleftharpoons \text{SO}_2(g) + \frac{1}{2}\ \text{O}_2(g)$$
When a chemical reaction is reversed, its equilibrium constant is the reciprocal of the original:
$$K_{\text{reversed}} = \frac{1}{K_1}$$
* Step B: Multiply the coefficients. To match the stoichiometry of Reaction 2, we must multiply all coefficients of the reversed reaction by a factor of 2:
$$2\ \text{SO}_3(g) \rightleftharpoons 2\ \text{SO}_2(g) + \text{O}_2(g)$$
When the stoichiometric coefficients of a reaction are multiplied by a factor ($n$), its equilibrium constant is raised to the power of $n$:
$$K_2 = (K_{\text{reversed}})^2 = \left(\frac{1}{K_1}\right)^2 = \frac{1}{(K_1)^2}$$

4. Verify the mathematical relationship algebraically:
Substitute the expression for $K_1$ into our derived relationship to confirm it mathematically simplifies to the expression for $K_2$:
$$\frac{1}{(K_1)^2} = \frac{1}{\left(\frac{[\text{SO}_3]}{[\text{SO}_2][\text{O}_2]^{1/2}}\right)^2} = \frac{1}{\frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}} = \frac{[\text{SO}_2]^2[\text{O}_2]}{[\text{SO}_3]^2} = K_2$$
This algebraic verification confirms that $K_2 = 1 / (K_1)^2$, identifying Option C as the correct choice.

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WHY_OTHERS_WRONG:

• Option A is incorrect: This relationship ($K_2 = (K_1)^2$) would only be correct if Reaction 2 was produced by multiplying Reaction 1 by 2 *without* reversing it (which would yield $2\ \text{SO}_2 + \text{O}_2 \rightleftharpoons 2\ \text{SO}_3$). It fails to account for the reciprocal relationship required by reversing the reactants and products.
• Option B is incorrect: This suggests that both reactions are identical and share the same equilibrium constant. Since the reactants and products are reversed and the stoichiometric coefficients are different, their equilibrium constants cannot be equal.
• Option D is incorrect: This relationship ($K_2 = 1 / K_1$) only accounts for reversing the direction of Reaction 1. It fails to account for multiplying the reaction coefficients by 2, which requires squaring the reciprocal constant.