Homework 11.111

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Caitlin Mispagel 1D
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Joined: Tue Oct 10, 2017 7:13 am

Homework 11.111

Postby Caitlin Mispagel 1D » Fri Mar 16, 2018 6:46 pm

Can someone explain how to do number 111 in chapter 11? The question reads "A certain enzyme-catalyzed reaction in a biochemical cycle has an equilibrium constant that is 10 times the equilibrium constant of the next step in the cycle. If the standard Gibbs free energy of the 1st reaction is 200. kJ mol^-1, what is the standard Gibbs free energy of the second reaction?"

Andrea Grigsby 1I
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

Re: Homework 11.111

Postby Andrea Grigsby 1I » Fri Mar 16, 2018 7:31 pm

you would use the equation dG = - RT lnK for this question
rearrange and substitute in the values you have
lnK=- -200000/(8.314*298) = 80.7
solve for K
the question says the equilibrium constant of the first step is 10 times greater than that of the second step
so you simply divide the above K by 10 - in this case just minus 1 from the power to get
K=1.14*10^34 for the second reaction
then, substitute everything back into the same equation, this time solving for reaction 2
dG2= -8.314 * 298 * ln 1.14*10^34 = -194 kJ

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