Question

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

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] Rate (M) [H2] (M) (mol/L/s) 0.32 0.64 0.64 0.37 0.37 0.74 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.)

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

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