In a closed rigid vessel, the equilibrium partial pressures are: = 100 mm, = 400 mm and = 1000 mm. Now, nitrogen is removed from the vessel until the pressure of hydrogen at equilibrium is equal to 70

Question

In a closed rigid vessel, the equilibrium partial pressures are: $N_2$ = 100 mm, $H_2$ = 400 mm and $NH_3$ = 1000 mm. Now, nitrogen is removed from the vessel until the pressure of hydrogen at equilibrium is equal to 700 mm. The new equilibrium partial pressure of $N_2$ is:

Prashant Kotian · Accepted Answer

Correct answer: A. Step 1 - Write the equilibrium reaction and calculate initial Kp $$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$$ $$K_p = \frac{p_{\ce{NH3}}^2}{p_{\ce{N2}} \cdot p_{\ce{H2}}^3} = \frac{(1000)^2}{100 \times (400)^3} = \frac{10^6}{100 \times 6.4 \times 10^7} = \frac{10^6}{6.4 \times 10^9} = 1.5625 \times 10^{-4}\text{ mm}^{-2}$$ Step 2 - Analyze the backward shift using stoichiometry Removing N2 shifts equilibrium backward (Le Chatelier). H2 rises from 400 → 700 mm: $$p'_{\ce{H2}} = 400 + 3x = 700 \implies x = 100\text{ mm}$$ New ammonia pressure: $$p'_{\ce{NH3}} = 1000 - 2x = 1000 - 200 = 800\text{ mm}$$ Step 3 - Solve for new N2 pressure using constant Kp Temperature is constant, so Kp is unchanged: $$K_p = \frac{(p'_{\ce{NH3}})^2}{p'_{\ce{N2}} \cdot (p'_{\ce{H2}})^3}$$ $$1.5625 \times 10^{-4} = \frac{(800)^2}{p'_{\ce{N2}} \times (700)^3} = \frac{640000}{p'_{\ce{N2}} \times 3.43 \times 10^8}$$ $$p'_{\ce{N2}} = \frac{640000}{1.5625 \times 10^{-4} \times 3.43 \times 10^8} = \frac{640000}{53593.75} \approx \boxed{11.94\text{ mm}}$$ Step 4 - Evaluate all options - **Option (A) 11.94 mm**: Correct. Backward shift gives P_NH3=800, P_H2=700, Kp gives P_N2'=11.94 mm. - **Option (B) 200 mm**: Incorrect. Does not satisfy the equilibrium constant. - **Option (C) 18.66 mm**: Incorrect. Computational error in applying Kp with new pressures. - **Option (D) 43.78 mm**: Incorrect. Mathematical error.

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