In: Chemistry
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?
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