The rate of a certain reaction is given by the following rate law: 2 rate-k [N2]

Question

The rate of a certain reaction is given by the following rate law: 2 rate-k [N2]2[ rate-k NH Use this information to answer the questions below. What is the reaction order in N2? x10 What is the reaction order in H2? What is overall reaction order? At a certain concentration of N2 and H2, the initial rate of reaction is 5.0 x 104 M /s. What would the initial rate of the reaction be if the concentration of N2 were doubled? Round your answer to 2 significant digits. The rate of the reaction is measured to be 6.0 M / s when [N2] = 0.77 M and [H2] = 1.7 M. Calculate the value of the rate constant. Round your answer to 2 significant digits.

Explanation / Answer

conisder the rate law

rate = k [N2]^2 [H2]^2

so

reaction order in N2 ======> 2

reaction order in H2 =====> 2

overall reaction order ====> 2 + 2 = 4

now

rate = k [N2]^2 [H2]^2

so

if the concentration of N2 were doubled , the rate will be increased by 4 times

so

new initial rate = 4 x 5 x 10^4

new initial rate = 2 x 10^5 M/s

3)

6 = k x [0.77]^2 [1.7]^2

k = 3.5

so

rate constant k is 3.5 M-3 s-1

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