A student obtained the following data for the rearrangement of cyclopropane to propene at 500 °C. (CH2)3(g)CH3CH=CH2(g) [(CH2)3], M 0.128 6.40×10-2 3.20×10-2 1.60×10-2 time, min 0 14.4 28.8 43.2 (1) What is the half-life for the reaction starting at t=0 min? min What is the half-life for the reaction starting at t=14.4 min? min Does the half-life increase, decrease or remain constant as the reaction proceeds? (2) Is the reaction zero, first, or second order? (3) Based on these data, what is the rate constant for the reaction? min-1

Explanation:

CH2)3(g)CH3CH=CH2(g) [(CH2)3], M       time, min

0.128               0

6.40×10-2          14.4

3.20×10-2        28.8

1.60×10-2          43.2

(1) What is the half-life for the reaction starting at t=0 min? min

Half life is the amount of time required for a substance to decay by half of it's initial concentration.

Starting form 0, the initial concentration = 0.128

After 14.4 mins, the final concentration is now exactly half of the initial concentration. This means 14.4 min is the half life starting from t=0min

What is the half-life for the reaction starting at t=14.4 min?

Starting form 14.4min, the initial concentration = 6.40×10-2

After 14.4 mins (28.8 - 14.4), the final concentration is now exactly half of the initial concentration. This means 14.4 min is the half life starting from t=14.4min

Does the half-life increase, decrease or remain constant as the reaction proceeds?

The half life is a constant factor, hence it remains constant as the reaction proceeds.

(2) Is the reaction zero, first, or second order?

Because the half life is independent of the concentration, it is a first order reaction.

In a zero order reaction, the half life Decreases as the reaction progresses; as concentration decreases.

In a first order reaction, the half life Increases with decreasing concentration.

(3) Based on these data, what is the rate constant for the reaction? min-1

The realtionship between the half life and rate onstant is;

k = 0.693 / half life

k = 0.693 / 14.4

k = 0.048125 min-1

the answer would be b i believe