Question

4.5 moles each of hydrogen and iodine is heated in a sealed ten litre vessel. At equilibrium, 3 moles of HI were found. The equilibrium constant for H₂(g) + I₂(g) ⇋ 2HI(g) is _____________ .

Answer 1

Correct Answer: 1

To calculate the equilibrium constant \( K_c \) for the reaction H₂(g) + I₂(g) ⇋ 2HI(g), follow these steps:
1. **Initial Concentrations:**
Initial moles of H₂ = 4.5;
Initial moles of I₂ = 4.5;
Volume of the vessel = 10 L;
Thus, initial concentrations are:
[H₂]_initial = [I₂]_initial = \(\frac{4.5}{10} = 0.45 \, \text{M}\).
2. **Change in Concentrations:**
At equilibrium, 3 moles of HI are found. Thus, [HI]_equilibrium = \(\frac{3}{10} = 0.3 \, \text{M}\).
According to the reaction stoichiometry, the formation of 2 moles of HI corresponds to the consumption of 1 mole each of H₂ and I₂. Therefore, the change for H₂ and I₂ is \(-\frac{3}{2 \times 10} = -0.15 \, \text{M}\).
3. **Equilibrium Concentrations:**
[H₂]_equilibrium = 0.45 - 0.15 = 0.3 M;
[I₂]_equilibrium = 0.45 - 0.15 = 0.3 M.
4. **Expression for \( K_c \):**
The equilibrium constant expression is given by:
\( K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.3)^2}{0.3 \cdot 0.3} = \frac{0.09}{0.09} = 1 \).
5. **Verification:**
The calculated \( K_c = 1 \) falls within the provided range of [1, 1].
Thus, the equilibrium constant \( K_c \) is confirmed to be 1, fitting perfectly within the given range.