## 15.63

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

Grace Boyd 2F
Posts: 50
Joined: Thu Jul 27, 2017 3:01 am

### 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 Celsius. Determine the rate constant at blood temperature (37 degrees C) given that the activation energy for the reaction is 38 kJ/mol.

Could someone please explain step by step how to do this problem? Thank you!

Sarah_Stay_1D
Posts: 57
Joined: Sat Jul 22, 2017 3:00 am
Been upvoted: 1 time

### Re: 15.63

Grace Boyd 2F wrote: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 Celsius. Determine the rate constant at blood temperature (37 degrees C) given that the activation energy for the reaction is 38 kJ/mol.

Could someone please explain step by step how to do this problem? Thank you!

We haven't learned this equation in class yet but for this problem you would use the equation ln(k2/k1)=(-Ea/R)(1/T1 - 1/T2). All you need to do for this problem is plug everything in and solve.
k1=1.5x10^10 L/mol s
Ea=38 kJ/mol
T1= 25 degrees C + 273 = 298 k
T2 = 37 degrees C + 273 = 310 k
R (gas constant) = 8.314 J/K mol = 8.314x10^-3 kJ/K mol
Then you just solve for k2.

Posts: 88
Joined: Fri Sep 29, 2017 7:03 am

### Re: 15.63

For that equation, do you know if it would matter if we switched k and k' as long as the corresponding temperatures are the same? Would you end up getting the same result?

Posts: 88
Joined: Fri Sep 29, 2017 7:03 am

### Re: 15.63

Also, for that question, do you know where the 1.8 from k'/k comes from and the -0.59 comes from at the end? When I plug in my numbers I end up getting something like 1.06 for k'/k, so my numbers are off.

Seth_Evasco1L
Posts: 54
Joined: Thu Jul 27, 2017 3:00 am

### Re: 15.63

I believe there are several typos in 15.63 of the solution manual:

The first typo being that 0.08314 kJ*K-1*mol-1 should really be 0.008314 kJ*K-1*mol-1 when converting the rate constant R in terms of kJ.

When you use that to calculate ln(k'/k), you should get approximately 0.59 which is where I think the second typo comes in? I think when they plug in those numbers, that "-0.59" should be an "=0.59" instead.

You take the natural log of both sides which gives k'/k = 1.8 and the rest of the problem should be the same as what is written in the solution manual.

Seth_Evasco1L
Posts: 54
Joined: Thu Jul 27, 2017 3:00 am

### Re: 15.63

Another typo is in the initial statement at the top of the solution given. The temperature in Kelvin should actually read 310 K and not 3130 K.