Question

The reaction, A + 2 B → C was studied. The reagents A and B were mixed and the time interval until a certain quantity of product C accumulated was measured.

run 1: [A]_initial = 0.100, [B]_initial = 0.140, time = 25 s

run 2: [A]_initial = 0.050, [B]_initial = 0.140, time = 50 s

run 3: [A]_initial = 0.100, [B]_initial = 0.070, time = 100 s

From this data, one conclusion that can be made is that the reaction is:

Select one:

a. first order with respect to substance A

b. zero order with respect to substance A

c. one-half order with respect to substance A

d. second order with respect to substance A

e. third order with respect to substance B

Answer 1

Let the rate of the reaction be denoted as

Rate = k*[A]^m[B]ⁿ

Consider runs 1 and 2. The concentration of B is kept fixed, while the concentration of A is halved.

Assume the final concentration falls to zero. Therefore,

Rate of disappearance of A = -[(concentration at time t₂) – (concentration at time t₁)]/(t₂- t₁)

For run 1, we have,

Rate = -[(0.00 M) – (0.100 M)]/(25 – 0) s = -(-0.100 M/25 s) = 0.004 M/s

For run 2, we have,

Rate = -[(0.00 M) – (0.050 M)]/(50 – 0) s = -(-0.050 M/50 s) = 0.001 M/s

Therefore,

(Rate)₁/(Rate)₂ = (0.004 M/s)/(0.001 M/s) = k*(0.100)^m(0.140)ⁿ/k.(0.050)^m(0.140)ⁿ

===> 4 = (2)^m

===> 2² = 2^m

===> m = 2

Next consider runs 1 and 3.

Rate of disappearance of B = -[(concentration at time t₂) – (concentration at time t₁)]/(t₂- t₁)

For run 1, we have,

Rate = -[(0.00 M) – (0.140 M)]/(25 – 0) s = -(-0.140 M/25 s) = 0.0056 M/s

For run 3, we have,

Rate = -[(0.00 M) – (0.070 M)]/(100 – 0) s = -(-0.070 M/100 s) = 0.0007 M/s

Therefore,

(Rate)₁/(Rate)₃ = (0.0056 M/s)/(0.0007 M/s) = k*(0.100)^m(0.140)ⁿ/k.(0.100)^m(0.070)ⁿ

===> 8 = (2)ⁿ

===> 2³ = 2ⁿ

===> n = 3

Therefore, the reaction is 2^nd order in A and 3^rd order in B. Therefore, both (d) and (e) are correct options.