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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

Sagot :

jufsanabriasa

Answer:

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 =

4.79x10¹³s⁻¹

s⁻¹ because the reaction is first order.

pstnonsonjoku

The pre - exponential factor is obtained form the information given ion the question as  4.79 × 10^13 s−1.

For a first order reaction, the Arrhenius equation is written as;

lnk = (-Ea/R)1/T + lnA. This is similar to the equation of a straight line, y = mx + c.

The pre-exponential factor is the intercept (c). Hence;

ln A = 31.5

A = e^31.5

A = 4.79 × 10^13 s−1

Learn more: https://brainly.com/question/20906233

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