H2(g) + I2(g) <-> 2 HI(g) Kc = 50. at 600°C Equimolar samples of each of three gases, H2(g), I2(g), — Equilibrium Chemistry Question

STEPS:

1. Write the expression for the reaction quotient ($Q_c$):
For the given gas-phase equilibrium system:
$$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\ \text{HI}(g)$$
The reaction quotient, which measures the relative amounts of products and reactants at any given point in time, is written as:
$$Q_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$$

2. Define the initial concentrations of the gases:
* The problem states that equimolar samples of $\text{H}_2(g)$, $\text{I}_2(g)$, and $\text{HI}(g)$ are introduced into a $3.0\text{ L}$ container.
* "Equimolar" means that the same number of moles ($n$) of each gas is initially present. Since all three gases occupy the same volume, their initial molar concentrations must be identical:
$$[\text{H}_2]_0 = [\text{I}_2]_0 = [\text{HI}]_0 = x\text{ M}$$

3. Calculate the initial value of $Q_c$:
* Substitute the identical initial concentrations ($x$) into the reaction quotient expression:
$$Q_c = \frac{(x)^2}{(x)(x)} = \frac{x^2}{x^2} = 1$$
* Because the concentration terms cancel out mathematically, the initial value of $Q_c$ is exactly 1, regardless of the specific number of moles added or the volume of the vessel.

4. Compare $Q_c$ to the equilibrium constant ($K_c$):
* We are given that the equilibrium constant is $K_c = 50.$ at $600^\circ\text{C}$.
* Comparing our calculated value to the equilibrium constant:
$$Q_c = 1 \quad \text{and} \quad K_c = 50. \implies Q_c < K_c$$

5. Determine the direction of the equilibrium shift:
* Because $Q_c < K_c$, the ratio of product concentration to reactant concentrations is smaller than it needs to be to achieve dynamic equilibrium.
* To reach equilibrium, the system must shift its position to the right (favoring the forward reaction) to convert reactants into products.
* This forward shift will consume the reactants ($\text{H}_2$ and $\text{I}_2$) and produce more of the product ($\text{HI}$) until the reaction quotient rises to equal the equilibrium constant of 50. This matches Option C.

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WHY_OTHERS_WRONG:

• Option A is incorrect: This option claims that $Q_c > K_c$. This is mathematically incorrect because $1 < 50$. If $Q_c$ were indeed greater than $K_c$, the reaction would shift to the left to produce more reactants like $\text{H}_2$, but that is not the case here.
• Option B is incorrect: While this option correctly states that $Q_c < K_c$, it makes a stoichiometric error in the predicted shift. When $Q_c < K_c$, the forward reaction is favored, which *consumes* reactant $\text{I}_2(g)$ rather than producing more of it.
• Option D is incorrect: This option falsely assumes the system is already in dynamic equilibrium ($Q_c = K_c$). Since $Q_c$ (1) does not equal $K_c$ (50), a net chemical change must occur to establish equilibrium.