Ge(g) + 2 Cl2(g) ⇄ GeCl4(g) The value of the equilibrium constant for the reaction represented above — Equilibrium Chemistry Question

STEPS:

1. Analyze the relationship between the initial and target chemical equations:
* Given equation: $\text{Ge}(g) + 2\ \text{Cl}_2(g) \rightleftharpoons \text{GeCl}_4(g)$ with $K_1 = 1 \times 10^{10}$
* Target equation: $2\ \text{GeCl}_4(g) \rightleftharpoons 2\ \text{Ge}(g) + 4\ \text{Cl}_2(g)$ with $K_2 = ?$

2. Apply the rule for reversing a reaction:
* To place $\text{GeCl}_4(g)$ on the reactant side, we must reverse the given equation:
$$\text{GeCl}_4(g) \rightleftharpoons \text{Ge}(g) + 2\ \text{Cl}_2(g)$$
* When a chemical equation is reversed, its equilibrium constant is the reciprocal of the original value:
$$K_{\text{reversed}} = \frac{1}{K_1} = \frac{1}{1 \times 10^{10}} = 1 \times 10^{-10}$$

3. Apply the rule for multiplying stoichiometric coefficients:
* To obtain the final target equation, we must multiply all coefficients by 2:
$$2\ \text{GeCl}_4(g) \rightleftharpoons 2\ \text{Ge}(g) + 4\ \text{Cl}_2(g)$$
* When the stoichiometric coefficients of a reaction are multiplied by a factor ($n$), the equilibrium constant is raised to the power of $n$:
$$K_2 = (K_{\text{reversed}})^2 = \left(\frac{1}{K_1}\right)^2$$

4. Calculate the final numerical value ($K_2$):
* Substitute the values into the mathematical relationship:
$$K_2 = (1 \times 10^{-10})^2 = 1 \times 10^{-20}$$
* This calculation confirms that Option A is the correct answer.

*

WHY_OTHERS_WRONG:

• Option B is incorrect ($1 \times 10^{-10}$): A student would arrive at this value if they recognized that the reaction was reversed (taking the reciprocal of $K_1$) but forgot to account for multiplying the coefficients by 2 (neglecting to square the reciprocal constant).
• Option C is incorrect ($1 \times 10^{10}$): This choice incorrectly assumes that reversing and changing the stoichiometry of the reaction has no effect on the value of the equilibrium constant.
• Option D is incorrect ($1 \times 10^{20}$): A student would get this value if they correctly squared the equilibrium constant to account for doubling the coefficients ($(1 \times 10^{10})^2 = 1 \times 10^{20}$) but forgot to invert the constant to account for reversing the direction of the reaction.