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.
|Rate (mol/L/s)||3.410 ✕ 10−3||1.364 ✕ 10−2||2.728 ✕ 10−2|
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
rate = k [H2][NO]2
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