At 300 K, Kc=1.65x10-10 for the reaction: N2O(g) + NO2(g) ⇄ 3NO(g) If 0.400 mol of...

At 300 K, Kc=1.65x10-10 for the reaction: N2O(g) + NO2(g) ⇄ 3NO(g) If 0.400 mol of N2O and 0.600 mol NO2 are added into a 4.00 L container, what will the concentration NO be at equilibrium?

Expert Solution

Equilibrium reaction: N2O(g)   +   NO2(g)   <--->    3NO(g)

We have 0.4 mol of N2O, and 0.6 mol of NO2 ...so we need to convert them into molarities before using the ICE table;

[N2O] = 0.4 mol/4.0 L = 0.1 M
[NO2] = 0.6 mol/4.0 L = 0.15 M

                   N2O(g)   +   NO2(g)   <--->    3NO(g)
Initial               0.1 M            0.15 M            0
Change               -x               -x               3x
Equilibrium           0.1-x            0.15-x           3x

Kc = [NO]^3/[N2O][NO2]
1.65x10^-10= (3x)^3/(0.1-x)(0.15-x)
(Since the K is very small, the value of (0.1 - x) ~ 0.1 and (0.15-x) ~ 0.15 )

Therefore,
1.65x10^-10= (3x)^3/(0.1)(0.15)
2.475 x 10^-12 = 9 x^3
x^3 = 0.275 x 10^-12
x = 0.65 x 10^-4

[NO] = 3x = 3(0.65 x 10^-4 ) = 1.95 x 10^-4 M

Hence, the concentration of [NO] at equilibrium is 1.95 x 10^-4 M

queen_honey_blossom answered 3 years ago

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