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Lucia H 2L
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Joined: Mon Nov 14, 2016 3:00 am
Been upvoted: 1 time


Postby Lucia H 2L » Sat Feb 10, 2018 5:28 pm

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 rst reaction is 200. kJ mol 1, what is the standard Gibbs free energy of the second reaction?

So k2 = k1/10
using delta G = -RTlnk, k1 = e^(200/RT)
plugging in the value for k2, delta G = -RTln(1/10(e^200/RT))
How do you calculate this without a given temp?

Andres Reynoso 1J
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

Re: 11.111

Postby Andres Reynoso 1J » Sat Feb 10, 2018 6:54 pm

Because the question asks for the standard Gibbs free energy of the second reaction, we assume that the temperature the reaction occurred at is 25°C since that is the accepted temperature for standard conditions.

Adrian Lim 1G
Posts: 88
Joined: Fri Sep 29, 2017 7:03 am

Re: 11.111

Postby Adrian Lim 1G » Mon Feb 12, 2018 1:48 am

Just as Andres stated, we assume the temperature is 25C because it is in standard state. However, when you plug T into the equations, make sure the temperature is in Kelvin because the unit of R is given in Kelvin

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