Question

Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the following equation.

H₂(g) + 2 NO(g) → N₂O(g) + H₂O(g)

Consider the below data.

[NO] (M)	0.32	0.64	0.64
[H2] (M)	0.37	0.37	0.74
Rate (mol/L/s)	3.410 ✕ 10−3	1.364 ✕ 10−2	2.728 ✕ 10−2

(a)

Determine the rate equation from the given data. (Rate expressions take the general form: rate = k . [A]^a . [B]^b.)

Answer 1

H2(g) + 2NO(g) ------>N2O(g) + H2O(g)

Rate = k [H2]^a[NO]^b

Compare the first experiment with second experiment, in second experiment concentration of NO is doubled and rate of the reaction increased four fold. So,

rate = k [H2]^a(2×[NO])^b

rate = k(2)^b [H2]^a[NO]^b

(2)^b = 4

b = 2

Compare the second experiment with third experiment, concentration of H2 is doubled and rate of the reaction is doubled. So,

rate = k (2×[H2])^a[NO]^b

rate = k (2)^a[H2]^a[NO]^b

(2)^a = 2

a = 1

Therefore,

rate = k [H2][NO]²