Question

From the following data for the reaction between A and B:

[A]	[B]	Initial rate (mol L−1 s-1) at
		300 K	320 K
2.5 × 10−4	3.0 × 10−5	5.0 × 10−4	2.0 × 10−3
5.0 × 10−4	6.0 × 10−5	4.0 × 10−3	-
1.0 × 10−3	6.0 × 10−5	1.6 × 10−2	-

Calculate

1. the order of the reaction with respect to A and with respect to B,
2. the rate constant at 300K,
3. the energy of activation, and
4. the pre-exponential factor.

Answer 1

i. Suppose the order of the reaction with respect to A is p and that with respect to B is q. Therefore, the rate law is

Rate = k [A]^p [B]^q.

At 300 K, we have

5.0 × 10⁻⁴ = k³⁰⁰ [2.5 × 10⁻⁴]^p [3.0 × 10⁻⁵]^q    ...(i)

4.0 × 10⁻³ = k³⁰⁰ [5.0 × 10⁻⁴]^p [6.0 × 10⁻⁵]^q    ...(ii)

1.6 × 10⁻² = k³⁰⁰ [1.0 × 10⁻³]^p [6.0 × 10⁻⁵]^q    ...(iii)

Dividing eq. (ii) by eq. (iii), we get

`(4.0 xx 10^-3)/(1.6 xx 10^-2) = ((5.0 xx 10^-4)/(1.0 xx 10^-3))^p`

or, `1/4 = (1/2)^p`

or, `(1/2)^2 = (1/2)^p`

⇒ p = 2

Dividing eq. (i) by eq. (ii) and putting the value of p, we get

`(5 xx 10^-4)/(4.0 xx 10^-3) = ((2.5 xx 10^-4)/(5.0 xx 10^-4))^2 ((3.0 xx 10^-5)/(6.0 xx 10^-5))^q`

or, `1/8 = (1/2)^2 xx (1/2)^q`

or, `(1/2)^1 = (1/2)^q`

∴ q = 1

Hence, the given reaction is of order 2 with respect to A and of order 1 with respect to 8.

ii. Putting the values of p and q in eq. (i), we have

5.0 × 10⁻⁴ = k³⁰⁰ (2.5 × 10⁻⁴)² (3.0 × 10⁻⁵)¹

∴ k³⁰⁰ = `(5.0 xx 10^-4)/((2.5 xx 10^-4)^2 xx (3.0 xx 10^-5))`

= `0.0005/(0.0000000625 xx 0.00003`

= `0.0005/0.000000000001875`

= 2.67 × 10⁸ s⁻¹.

iii. At 320 K,

2.0 × 10⁻³ = k³²⁰ × (2.5 × 10⁻⁴)² × (3.0 × 10⁻⁵)¹

∴ k³²⁰ = `(2.0 xx 10^-3)/((2.5 xx 10^-4)^2 xx (3.0 xx 10^-5))`

= 1.07 × 10⁹ s⁻¹.

According to the Arrhenius equation,

`log_10  k_2/k_1 = E_a/(2.303 R) [1/T_1 - 1/T_2]`

For T₁ = 300 K, and T₂ = 320 K, we get

`log_10  (1.07 xx 10^9)/(2.67 xx 10^8) = E_a/(2.303 xx 8.314) [1/300 - 1/320]`

or, E_a = 55407.8 J mol⁻¹ = 55.41 kJ mol⁻¹.

iv. According to the Arrhenius equation,

k = `A e^(-E_a//RT)`

∴ log_e k = `log_e A - E_a/(RT)`

or, log₁₀ k = `log_10 A - E_a/(2.303 RT)` 

At 300 K,

log₁₀ 2.67 × 10⁸ = `log_10 A - 55407.8/(2.303 xx 8.314 xx 300)`

or, 8.4265 = log₁₀ A − 9.6460

or, A = antilog₁₀ (8.4265 + 9.6460)

= 1.18 × 10¹⁸ s⁻¹.