X(g) + 2 Q(g) ⇌ R(g) + Z(g) K_c = 1.3 × 10^5 at 50°C A 1.0 mol sample of X(g) and a 1.0 mol sample o — Equilibrium Chemistry Question

STEPS:

1. Calculate the initial concentrations of the reactants:
* The container is an evacuated, rigid $10.0\text{ L}$ vessel.
* The initial amounts of reactants are $1.0\text{ mol}$ of $\text{X}(g)$ and $1.0\text{ mol}$ of $\text{Q}(g)$.
* Using the molarity formula ($M = \frac{n}{V}$):
$$[\text{X}]_0 = \frac{1.0\text{ mol}}{10.0\text{ L}} = \mathbf{0.10\text{ M}}$$
$$[\text{Q}]_0 = \frac{1.0\text{ mol}}{10.0\text{ L}} = \mathbf{0.10\text{ M}}$$
$$[\text{R}]_0 = [\text{Z}]_0 = \mathbf{0\text{ M}}$$

2. Analyze the equilibrium constant ($K_c$):
* The equilibrium constant is $K_c = 1.3 \times 10^5$ at $50^\circ\text{C}$.
* A very large equilibrium constant ($K_c \gg 1$) indicates that the forward reaction is highly thermodynamically favorable and will proceed almost to completion.

3. Identify the limiting reactant:
* According to the balanced chemical equation:
$$\text{X}(g) + 2\ \text{Q}(g) \rightleftharpoons \text{R}(g) + \text{Z}(g) \quad \text{}$$
* The stoichiometric ratio requires $2\text{ moles}$ of $\text{Q}$ to completely react with $1\text{ mole}$ of $\text{X}$.
* Since we only have $1.0\text{ mole}$ of each reactant, $\text{Q}$ is the limiting reactant because completely consuming $1.0\text{ mol}$ of $\text{X}$ would require $2.0\text{ mol}$ of $\text{Q}$.

4. Estimate the equilibrium concentrations using stoichiometry:
* Let the change in concentration of the products be $+x$. The ICE table can be set up as:
* $[\text{X}]_{\text{eq}} = 0.10 - x$
* $[\text{Q}]_{\text{eq}} = 0.10 - 2x$
* $[\text{R}]_{\text{eq}} = x$
* $[\text{Z}]_{\text{eq}} = x$
* Since the reaction goes nearly to completion, the limiting reactant $\text{Q}$ is almost entirely consumed:
$$0.10 - 2x \approx 0 \implies 2x \approx 0.10 \implies x \approx \mathbf{0.050\text{ M}}$$
* Substitute $x \approx 0.050\text{ M}$ back to estimate the equilibrium concentrations:
* $[\text{R}]_{\text{eq}} \approx \mathbf{0.050\text{ M}}$
* $[\text{Z}]_{\text{eq}} \approx \mathbf{0.050\text{ M}}$
* $[\text{X}]_{\text{eq}} \approx 0.10 - 0.050 = \mathbf{0.050\text{ M}}$
* $[\text{Q}]_{\text{eq}} \approx \mathbf{\approx 0\text{ M}}$ (a very small, non-zero concentration remains due to equilibrium)

5. Compare the final concentrations:
* Because both $\text{R}$ and $\text{Z}$ are produced in a $1:1$ stoichiometric ratio starting from $0\text{ M}$, their concentrations must be exactly equal at equilibrium: $[\text{R}] = [\text{Z}] \approx 0.050\text{ M}$.
* The concentration of the limiting reactant $\text{Q}$ is nearly $0\text{ M}$.
* Therefore, $[\text{R}] = [\text{Z}] > [\text{Q}]$, which perfectly matches Option C.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: In the original exam formatting, this option was $[\text{R}] = \frac{1}{2}[\text{Q}]$ (fractured by scanning as `1 [R] = [Q] 2`). This is incorrect because $[\text{R}] \approx 0.050\text{ M}$ while $[\text{Q}] \approx 0\text{ M}$. If interpreted as the typed $[\text{R}] > [\text{Q}]$, while true, it is incomplete compared to Option C because it fails to establish the stoichiometric equality between the products $\text{R}$ and $\text{Z}$.
• Option B is incorrect: In the original exam formatting, this option was $[\text{Q}] = \frac{1}{2}[\text{X}]$ (scanned as `[Q] = [X] 2 1`). This is incorrect because $[\text{Q}]$ is nearly consumed ($\approx 0\text{ M}$) while $[\text{X}] \approx 0.050\text{ M}$. If interpreted as the typed $[\text{Q}] < [\text{X}]$, it is also incomplete as a description of the final equilibrium state.
• Option D is incorrect: This option assumes that the concentrations of all reactants and products are equal at equilibrium. This only occurs if $K_c$ is close to 1 and the starting species have stoichiometric amounts that do not result in a limiting reactant, which is not the case for this reaction.