Question

For the nitrogen fixation reaction, 3h2(g) + n2(g) 2nh3(g), kc = 6.0 × 10–2 at 500°c. if 0.250 m h2 and 0.050 m nh3 are present at equilibrium, what is the equilibrium concentration of n2?a. 3.3 m
b. 2.7 m
c. 0.20 m
d. 0.083 m
e. 0.058 m

Answer 1

Answer:

b. 2.7 M

Explanation:

Hello,

In this case, by considering the law of mass action for the studied reaction:

$Kc=\frac{[NH_3]^2_{eq}}{[H_2]^3_{eq}[N_2]_{eq}}$

Now, since the concentration $N_2$ is the unique unknown, solving for it one obtains:

$[N_2]_{eq}=\frac{[NH_3]^2_{eq}}{[H_2]^3_{eq}Kc}$

$[N_2]_{eq}=\frac{(0.050M)^2}{(0.250M)^3(6.0x10^{-2})}$

$[N_2]_{eq}=2.7M$

Hence the answer is b.

Best regards.

Answer 2

Write the Kc equation...
Substitute the given values in the equation and u qill ontain the answer... Give it a try