The rate of a certain reaction is given by the following rate law: rate = k[H_2]

Question

The rate of a certain reaction is given by the following rate law: rate = k[H_2][I_2 Use this information to answer the questions below. What is the reaction order in H_2? What is the reaction order in I_2? What is overall reaction order? At a certain concentration of H_2 and I_2, the initial rate of reaction is 2.0 times 10^4 M/s. What would the initial rate of the reaction be if the concentration of H_2 were doubled? Round your answer to 2 significant digits. The rate of the reaction is measured to be 29.0 M/s when [H_2] = 1.9 M and [I_2] = 1.1 M. Calculate the value of the rate constant. Round your answer to 2 significant digits.

Explanation / Answer

Rate = k[H2][I2]

a).Reaction order in H2 = 1

b).Reaction order in I2 =1

c).Overall reaction order = 1+1 =2

d).Initial rate of reaction = 2.0×104M/s

So, 2.0 × 104 = k[H2][I2]

If H2 was doubled,rate would be

Rate = k[2H2][I2]

Rate = 2k[H2][I2]

Rate = 2 × 2.0 × 104 M/s

Initial rate of reaction = 4.0 × 10-4 M/s

e).Rate = k[H2][I2]

k = Rate/[H2][I2]

From values given

k = 29.0Msec-1/1.9M×1.1M

k = 29.0Msec-1/2.09M2

k = 14 /Msec

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