For the reaction N₂ + 3H₂ ⇌ 2NH₃, Kc = 0.5 at 400°C. If [N₂] = 0.1 M and [H₂] = 0.3 M at equilibrium, find [NH₃].

Kc = ([NH₃]²)/([N₂][H₂]³) 0.5 = ([NH₃]²)/(0.1 × (0.3)³) = ([NH₃]²)/(0.1 × 0.027) = ([NH₃]²)/(0.0027) [NH₃]² = 0.5 × 0.0027 = 1.35×10⁻³ [NH₃] = 0.0368 M

JEE chemistry

For the reaction N₂ + 3H₂ ⇌ 2NH₃, Kc = 0.5 at 400°C — JEE Chemistry

RJRahul Joshi · 12 Asked 1mo ago 916 views 1 answer

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For the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$, $K_c = 0.5$ at 400°C. If $[N_2] = 0.1$ M and $[H_2] = 0.3$ M at equilibrium, find $[NH_3]$.

1 Answer

PPrakashGurung52 ✓ Accepted · 1mo ago ▲ 18

$$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$$

$$0.5 = \frac{[NH_3]^2}{0.1 \times (0.3)^3} = \frac{[NH_3]^2}{0.1 \times 0.027} = \frac{[NH_3]^2}{0.0027}$$

$$[NH_3]^2 = 0.5 \times 0.0027 = 1.35\times10^{-3}$$

$$[NH_3] = 0.0368 M$$

Discussion (4)

How do we know the approximation is valid here?

Isha Malhotra · 1mo ago

Quick doubt: would this method still work if the numbers were not so clean?

Vikram Nair · 1mo ago

How do we know the approximation is valid here?

Divya Mehta · 1mo ago

What changes if the medium/conditions were different?

Kavya Krishnan · 1mo ago

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