Question

For a reaction with an activation energy of 52.0 kilojoules per mole and at a temperature of 35°C find the pre-exponential factor if the rate constant is 0.158 A) 0.161 В) 1.03 x 10% C) 6.42 x 1076 10 D) 2.42 x 10% E) 6.71 x 106

Answer 1

`A=1.03* 10^8`

Step-by-step explanation:

Using the expression,

`\ln {k}=ln\ A-(E_(a))/(RT)`

Wherem

`k\ is\ the\ rate\ constant\ at\ T`

`E_a` is the activation energy

A is the pre-exponential factor

R is Gas constant having value = 8.314 J / K mol

Thus, given that,
`E_a` = 52.0 kJ/mol = 52000 J/mol (As 1 kJ = 1000 J)

`k=0.158`

`T=35\ ^0C`

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (35 + 273.15) K = 308.15 K

Thus, applying values as:

`\ln {0.158}=ln\ A-(52000)/(8.314* 308.15)`

`A=e^{\ln \left(0.158\right)+(52000)/(8.314\cdot \:308.15)}`

`A=103161576.82851=1.03* 10^8`