15.61 (6th edition)
Posted: Wed Mar 13, 2019 11:18 pm
by Schuyler_Howell_4D
Can someone explain where they got the equation they use to solve for Ea:
ln(k'/k)=(Ea/R)[(1/T)-(1/T')]
Re: 15.61 (6th edition)
Posted: Thu Mar 14, 2019 12:28 am
by Andrea Zheng 1H
They got this by using the equation ln(k) = -Ea/(R*T). Because you are finding the difference in two different k values, you would use ln(k')-ln(k), which is ln(k'/k). On the other side of the equation, you would have Ea (which is constant) and R (which is constant) and T (which changes). This means you would end up with Ea/R multiplied by the difference in temperatures, which has 1/Temperature because temperature is in the denominator in ln(k) = -Ea/(R*T). Therefore, you get ln(k'/k) = (Ea/R)[(1/T)-(1/T')].