Question

The Kc for the following reaction at 225 ∘C is 1.7×102. 3H2(g)+N2(g)⇌2NH3(g)

If the equilibrium mixture contains 0.15 M H2 and 0.017 M N2, what is the molar concentration of NH3?

I have no idea what I am doing wrong here, help!

Answer 1

1) Chemical equation:

3H2 (g) + N2(g) = 2NH3 (g)

2) Equilibrium constant

                 [NH3]^2
=> Kc =  ------------------ = 1.7 * 10 ^2
               [H2]^3 * [N2]

3) Solve for the unknown

[NH3]^2 = 1.7 * 10^2 * [H]^3 * [N2]

[NH3]^2 = 1.7 * 10 ^2 * (0.15)^3 * (0.017) = 0.00975375

=> [NH3] = √(0.00975375) = 0.099

Answer: 0.099M

Answer 2

Answer:

Molar concentration of ammonia gas is 0.0987 M.

Explanation:

Concentration of hydrogen gas = $[H_2]=0.15 M$

Concentration of nitrogen gas = $[N_2]=0.017 M$

Concentration of ammonia gas = $[NH_3]=x$

Equilibrium constant of the reaction = $K_c=1.7\times 10^2$

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

$1.7\times 10^2=\frac{x^2}{(0.15 M)^3\times (0.017 M)}$

$9.5737\times 10^{-3} M^2=x^2$

x = 0.0987 M

Molar concentration of ammonia gas is 0.0987 M.