A plot of lnk against 1 T has a slope of−2575 K and an intercept of 31.5. What is the pre-exponential factor A(s−1) of this first order reaction

4.79x10¹³s⁻¹

Explanation:

For a first order reaction, the graph of ln k against 1/K follows Arrhenius equation:

lnK = ln A - Ea/RT

Where A is pre-exponential factor, Ea is activation energy, R gas constant and T is absolute temperature

If you graph ln k = y and 1/T = x:

y = lnA - Ea/R x

Where the slope = -Ea/R

And the intercept = lnA

To solve the pre-exponential factor:

31.5 = lnA

e^31.5 =

s⁻¹ because the reaction is first order.

answer: i would say b.

explanation: