Question

Use the rules for logarithms and exponents to write this equation in logarithmic form.?

Answer 1

Answer:

$ln K = ln (A) -\frac{\Delta H}{RT}$

Step-by-step explanation:

For this case we have the following expression:

$K = A e^{-\frac{\Delta H}{RT}}$   (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:

$ln K = ln(A e^{-\frac{\Delta H}{RT}})$

Using the following property:

$ln(xy) = ln (x) + ln(y)$ for x and y real numbers, x>0, y>0, then we have:

$ln K = ln (A) + ln (e^{-\frac{\Delta H}{RT}})$

Now since the natural log and the exponentiation are inverse operations we have this:

$ln K = ln (A) + (-\frac{\Delta H}{RT})$

And then the final expression for ln K is :

$ln K = ln (A) -\frac{\Delta H}{RT}$