CO(g) + 2 H2(g) ⇄ CH3OH(g) DH < 0 The synthesis of CH3OH(g) from CO(g) and H2(g) is represented by t — Equilibrium Chemistry Question

STEPS:

1. Calculate the initial concentrations of the reactants:
The rigid reaction vessel has a volume of $2.0\text{ L}$, into which $1.0\text{ mol}$ of $\text{CO}(g)$ and $1.0\text{ mol}$ of $\text{H}_2(g)$ are initially pumped. Using the molarity formula ($M = n/V$):
* $[\text{CO}]_0 = \frac{1.0\text{ mol}}{2.0\text{ L}} = \mathbf{0.50\text{ M}}$
* $[\text{H}_2]_0 = \frac{1.0\text{ mol}}{2.0\text{ L}} = \mathbf{0.50\text{ M}}$
* $[\text{CH}_3\text{OH}]_0 = \mathbf{0\text{ M}}$ (no product is initially present)
2. Determine the direction of the reaction shift to reach equilibrium:
Because the initial concentration of the product $\text{CH}_3\text{OH}$ is $0\text{ M}$, the initial reaction quotient ($Q_c$) is $0$. Since $Q_c < K_c$ (where $K_c = 14.5$), the reaction must shift in the forward direction to establish equilibrium. This means reactant molecules must be consumed to form product molecules.
3. Set up an ICE (Initial, Change, Equilibrium) table:
Let $x$ represent the change in concentration of $\text{CO}(g)$ as the system shifts forward. Because the reaction shifts forward to reach equilibrium, $x$ must be strictly positive ($x > 0$). Using the stoichiometry of the balanced equation:
* $\text{CO}(g)$: Initial = $0.50\text{ M}$; Change = $-x$; Equilibrium = $0.50 - x$
* $\text{H}_2(g)$: Initial = $0.50\text{ M}$; Change = $-2x$ (due to the 2:1 stoichiometric ratio); Equilibrium = $0.50 - 2x$
* $\text{CH}_3\text{OH}(g)$: Initial = $0\text{ M}$; Change = $+x$; Equilibrium = $x$
4. Compare the equilibrium concentrations of $\text{H}_2$ and $\text{CO}$:
Compare the algebraic expressions representing the concentrations of both reactants at equilibrium:
* $[\text{CO}]_{\text{eq}} = 0.50 - x$
* $[\text{H}_2]_{\text{eq}} = 0.50 - 2x$
Because $x > 0$ (as the reaction must proceed forward), subtracting $2x$ from $0.50$ will always yield a smaller value than subtracting only $x$ from $0.50$. Mathematically:
$$0.50 - 2x < 0.50 - x$$
$$[\text{H}_2]_{\text{eq}} < [\text{CO}]_{\text{eq}}$$
Therefore, at equilibrium, the concentration of $\text{H}_2$ is strictly less than the concentration of $\text{CO}$, which identifies Option B as the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect: For $[\text{H}_2] = 2[\text{CO}]$ to be true at equilibrium, we would need $0.50 - 2x = 2(0.50 - x)$, which simplifies to $0.50 - 2x = 1.00 - 2x$ and has no mathematical solution ($0.50 \neq 1.00$). Because both reactants start with equal concentrations ($0.50\text{ M}$) but $\text{H}_2$ is consumed twice as fast as $\text{CO}$, $\text{H}_2$ will always have a lower concentration than $\text{CO}$ at any point after the reaction begins.
• Option C is incorrect: At equilibrium, we have established that $[\text{H}_2] < [\text{CO}]$. Therefore, the inequality $[\text{CO}] < [\text{H}_2]$ presented in this option is factually incorrect. Additionally, for $[\text{CO}]$ to equal $[\text{CH}_3\text{OH}]$, we would need $0.50 - x = x$, which means $x = 0.25\text{ M}$. If $x = 0.25\text{ M}$, then $[\text{H}_2]_{\text{eq}} = 0.50 - 2(0.25) = 0\text{ M}$, which is impossible because in a reversible dynamic equilibrium, reactants can never be completely consumed.
• Option D is incorrect: For $[\text{CO}] = [\text{H}_2]$ to be true at equilibrium, we would need $0.50 - x = 0.50 - 2x$, which simplifies to $x = 0$. This would mean no reaction occurred at all, which is impossible because the system starts with zero product and must proceed forward to reach equilibrium.