Question:medium

For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium constant Kc is 0.5 at 400 K. If [N₂] = 0.1 M, [H₂] = 0.2 M, and [NH₃] = 0.05 M, what is the reaction quotient Qc?

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Carefully compute exponents for concentrations in equilibrium expressions to ensure accurate \( Q_c \) or \( K_c \) values.
Updated On: Jan 15, 2026
  • 0.3125
  • 0.625
  • 1.25
  • 2.5
Show Solution

The Correct Option is B

Solution and Explanation

To find the reaction quotient \( Q_c \):
  1. Define the \( Q_c \) expression: \[ Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}. \]
  2. Provided: \( [\text{N}_2] = 0.1 \, \text{M} \), \( [\text{H}_2] = 0.2 \, \text{M} \), \( [\text{NH}_3] = 0.05 \, \text{M} \).
  3. Calculate the numerator: \[ [\text{NH}_3]^2 = (0.05)^2 = 0.0025. \]
  4. Calculate the denominator: \[ [\text{H}_2]^3 = (0.2)^3 = 0.008, \] \[ [\text{N}_2] \times [\text{H}_2]^3 = 0.1 \times 0.008 = 0.0008. \]
  5. Compute \( Q_c \): \[ Q_c = \frac{0.0025}{0.0008} = \frac{2.5 \times 10^{-3}}{8 \times 10^{-4}} = \frac{2.5}{8} \times 10 = 0.625. \]
  6. Answer: 0.625 is option (B).
The answer: \[ \boxed{0.625} \]
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