Answer to Question #347450 in General Chemistry for Ashter

Question

A mixture of 0.100 mol of NO, 0.050 mol of H₂ and 0.10 mol of H₂O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established:

2NO(g) + 2H₂(g) ⇆ N₂(g) + 2H₂O(g).

At equilibrium [NO] = 0.062M. Calculate the equilibrium concentrations of H_2, N₂ and H₂O and Keq.

Answer 1

Initially,

[NO]= 0,1mol/l; [H₂]= 0,05mol/l; [H₂O]= 0,1mol/l

After chemical equilibrium [NO]= 0,062mol/l

we find how many reacts NO:

0,1-0,062 = 0,038

2NO + 2H₂ = N₂ + 2H₂O

0,038 — 0,038 — 0,019 — 0,038

[H₂]_equ = 0,05-0,038 = 0,012 mol/l

[N₂]_equ = 0,019 mol/l

[H₂O]_equ = 0,1+0,038 = 1,038 mol/l

[NO]_equ = 0,062 mol/l

$K_{equ}= \frac {[N_2][H_2O]}{[NO][H_2]}= \frac{0,019*1,038}{0,062*0,012}= 26,5$

Answer:

[H₂]_equ = 0,012 mol/l

[N₂]_equ = 0,019 mol/l

[H₂O]_equ = 1,038 mol/l

K_equ = 26,5