I don't get 5.04 for K but you're on the right track.
-4010 = 8.413*298*lnK
K = ? I obtained approx 0.2
Then
...........A + B ==> C
I.......0.30.0.40...0.0
C.........-x...-x....x
E.....0.30-x..0.40-x..x
Substitute the E line into Kc expression and solve for x, then evaluate 0.40-x and 0.30-x/
A reaction :
A(aq)+B(aq) --> C(aq)
has a standard free-energy change of –4.01 kJ/mol at 25 °C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively?
I used deltaG = -RT(lnK) to find that K= 5.04. Then I wasn't really sure what to do next. Please help.
2 answers
I think it might be because you used 8.413 for R, I think R is supposed to be 8.314. So I think K is right if it is 5.04.
But I seem to be stick on the substitution part because I tried to solve it but I get 5.04 = x/(0.3-x)(0.4-x). Am I supposed to multiply it out and isolate the variable? Because if I was supposed to I got 5.04x^2-4.528x+0.6048=0 and I tried to put it in the quadratic formula to get x=.81 or x =.00612 but they are both wrong
But I seem to be stick on the substitution part because I tried to solve it but I get 5.04 = x/(0.3-x)(0.4-x). Am I supposed to multiply it out and isolate the variable? Because if I was supposed to I got 5.04x^2-4.528x+0.6048=0 and I tried to put it in the quadratic formula to get x=.81 or x =.00612 but they are both wrong
1 answer