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 0.600 M/s. What would the initial rate of the reaction be if the concentration of Hp were doubled? The rate of the reaction is measured to be 55.0 M/s when [H_2] = 0.70 M and [I_2] = 0.80 M. Calculate the value of the rate constant. Round your answer to 2 significant digits.

Explanation / Answer

1)

Rate law is:

rate = k [H2] [I2]

power of H2 is 1

So, Order of H2 is 1

2)

Rate law is:

rate = k [H2] [I2]

power of I2 is 1

So, Order of I2 is 1

3)

overall order = sum of individual order

= 1 + 1

= 2

4)

Rate law is:

rate = k [H2] [I2]

Order of H2 is 1

Doubling concentration of H2 doubles the rate

SO,

rate becomes: 0.600*2 = 1.20 M/s

Answer: 1.20 M/s

5)

rate = k [H2] [I2]

55.0 M/s = k (0.70 M)*(0.80 M)

k = 98 M-1.s-1

Answer: 98 M-1.s-1

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