Question

The following concentrations were obtained for the formation of $NH_3(g)$ from $N_2(g)$ and $H_2(g)$ at equilibrium and at 500K: $[N_2] 1×10^{-2} M, [H_2] -2×10^{-2} M$ and $[NH_3]=2×10^{-2} M$. The equilibrium constant, K., for the reaction
$N_2(g)+3H_2(g)2NH; (g)$ at 500K is

• A. $5 \times10^{-3} mol^{-2}dm^6$
• B. $1 \times 10^3 mol^{-2}dm^6$
• C. $5 \times 10^{3} mol^{-2}dm^6$
• D. $ 2 \times 10^3 mol^{-2}dm^6$
• E. $2 \times 10^{-3} mol^{-2}dm^6$

Answer 1

The Correct Option is C

Equilibrium Constant for Ammonia Formation

Reaction: 
\( N_2(g) + 3H_2(g) ⇌ 2NH_3(g) \)

Given concentrations at equilibrium:

• \([N_2] = 1 \times 10^{-2} \, M\)
• \([H_2] = 2 \times 10^{-2} \, M\)
• \([NH_3] = 2 \times 10^{-2} \, M\)

Equilibrium constant expression:

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

Substituting the values:

\( K_c = \frac{(2 \times 10^{-2})^2}{(1 \times 10^{-2})(2 \times 10^{-2})^3} \)

Simplifying:

Numerator = \( 4 \times 10^{-4} \)
Denominator = \( 1 \times 10^{-2} \times 8 \times 10^{-6} = 8 \times 10^{-8} \)

\( K_c = \frac{4 \times 10^{-4}}{8 \times 10^{-8}} = 0.5 \times 10^4 = 5 \times 10^3 \, mol^{-2} \, dm^6 \)

Correct Answer: 5 × 10³ mol^-2dm⁶