Answer:
[tex]A=1.03\times 10^8[/tex]
Explanation:
Using the expression,
[tex]\ln {k}=ln\ A-\dfrac{E_{a}}{RT}[/tex]
Wherem
[tex]k\ is\ the\ rate\ constant\ at\ T[/tex]
[tex]E_a[/tex] is the activation energy
A is the pre-exponential factor
R is Gas constant having value = 8.314 J / K mol
Thus, given that, [tex]E_a[/tex] = 52.0 kJ/mol = 52000 J/mol (As 1 kJ = 1000 J)
[tex]k=0.158[/tex]
[tex]T=35\ ^0C[/tex]
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:
[tex]\ln {0.158}=ln\ A-\dfrac{52000}{8.314\times 308.15}[/tex]
[tex]A=e^{\ln \left(0.158\right)+\frac{52000}{8.314\cdot \:308.15}}[/tex]
[tex]A=103161576.82851=1.03\times 10^8[/tex]
