hw #15.63

$K = \frac{k_{forward}}{k_{reverse}}$

lauren chung 2f
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am

hw #15.63

The rate constant of the reaction between CO2 and OH- in aqueous solution to give the HCO3- ion is 1.5 x 10^10 L*mol-1*s-1 at 25 C. Determine the rate constant at blood temperature (37 C), given that the activation energy for the reaction is 38 kJ*mol-1.

I looked at the solution manual for help with this problem, but when I plug their numbers into my calculator, I keep getting 0.059 instead of 0.59 for ln(k1/k). Could someone help me figure out what I'm doing wrong?

Akash_Kapoor_1L
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

Re: hw #15.63

You would start with the equation: ln(k2/k11) = Ea/R * ((1/T1) - (1/T2))

Where T1 = 25 Degrees Celsius = 298 K
Where T2 = 37 Degrees Celsius = 310 K
Where k1 = 1.5 * 10^10 and you need to find k2

So after computing this equation and making sure that you convert the value for R from J into kJ, by making it 8.3145 * 10^-3
You will get the value for ln(k2/k11), you then find k2/k11 by putting that value (x) to the power of e -> so e^x
If you multiply that value by the value of k1, you should get the correct value of k2.