Answer:
[tex]ln K = ln (A) -\frac{\Delta H}{RT} [/tex]
Step-by-step explanation:
For this case we have the following expression:
[tex] K = A e^{-\frac{\Delta H}{RT}}[/tex] (1)
And we want to find the value of ln K. If we apply natural log on both sides of the equation (1) we got:
[tex] ln K = ln(A e^{-\frac{\Delta H}{RT}})[/tex]
Using the following property:
[tex] ln(xy) = ln (x) + ln(y)[/tex] for x and y real numbers, x>0, y>0, then we have:
[tex] ln K = ln (A) + ln (e^{-\frac{\Delta H}{RT}}) [/tex]
Now since the natural log and the exponentiation are inverse operations we have this:
[tex] ln K = ln (A) + (-\frac{\Delta H}{RT}) [/tex]
And then the final expression for ln K is :
[tex]ln K = ln (A) -\frac{\Delta H}{RT} [/tex]
