Consider the following reaction at 241 K: 1 A (aq) + 2 B (aq) 2 C (aq) + 1 D (aq
ID: 950616 • Letter: C
Question
Consider the following reaction at 241 K: 1 A (aq) + 2 B (aq) 2 C (aq) + 1 D (aq)
An experiment was performed with the following intitial concentrations: [A]i = 1.39 M, [B]i = 2.11 M, [C]i = 0.53 M, [D]i = 0.19 M. The reaction was allowed to proceed until equilibrium was reached at which time it was determined that [A] = 0.57 M. What was the maximum amount of work that could have been performed as the reaction began? Maximum amount of work performed as the reaction began?
the answer is (in kJ)= -16.8 kJ
but i dont know how to get this answer. can you show me how to get this answer?
Explanation / Answer
1 A + 2 B 2 C + 1 D
1.39__2.11__0.53__0.19
1.39-x_2.11-2x_0.53+2x_0.19+x
1.39-x=0.57M ----> x= 0.82M
K= [C]2[D]/[A][B]2 = (2.17)2(1.01)/(0.57)(0.47)2= 37.77
DG0= -RTLnK = -8.314x10-3kJ/mol.K x 241K x Ln37.77
DG0= -7.28
DG= DG0 + RTLnQ
Q is equal to Keq just with the initial values.
DG= -7.28 +RTLn(0.53)2(0.19)/(1.39)(2.11)2
DG= -16.8 kJ