I was working through the chapter 11 problem 11.111 in the book and was confused by the book answer. When the problem stated that the equilibrium constant for the forst reaction was 10 times more than the second reaction K value, the book ended up dividing by 100 in stead of 10. Was this a mistake or am I reading the problem wrong?
(I got 199.5 kj/mol instead of 194 kj/mol)
HW 11.111
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garrett_perozich
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Re: HW 11.111
Hi,
The book actually doesn't divide by 100. We know that K1 is 10x greater than K2. So that means that K1 = 10*K2. Now if we do K1/K1, substituting 10*K2 for K1, we get 10*K2/K2, or 10, since the K2's cancel. Now, that value is a constant for our equilibrium value when solving for our Gibbs Free Energy for step 2. Since we have that, we can set it equal to our equation for any arbitrary K value, which is e^(delta G / RT). But in this case, since this is a multi-step process, our delta G is going to be the sum of the Gibbs for each step. So now we can set 10 = e^( (G1+G2) / RT ) and solve for G2 (since we know G1).
Hope this helps!
The book actually doesn't divide by 100. We know that K1 is 10x greater than K2. So that means that K1 = 10*K2. Now if we do K1/K1, substituting 10*K2 for K1, we get 10*K2/K2, or 10, since the K2's cancel. Now, that value is a constant for our equilibrium value when solving for our Gibbs Free Energy for step 2. Since we have that, we can set it equal to our equation for any arbitrary K value, which is e^(delta G / RT). But in this case, since this is a multi-step process, our delta G is going to be the sum of the Gibbs for each step. So now we can set 10 = e^( (G1+G2) / RT ) and solve for G2 (since we know G1).
Hope this helps!
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