I can show you how.
dG = -RT*lnK
dG = -4.65 kJ/mol = 4650 J/mol
substitute in the above and solve for K. Then
............A + B ==> C
I..........0.3..0.4...0
C..........-x...-x....x
E........0.3-x...4-x..x
Substitute the E line into the Keq expression (use K from the first calculation with delta G) and solve for x. Then x, 0.3-x and 0.4-x will give you the equilibrium concns.
A reaction: A(aq) + B(aq) <-> C(aq)
has a standard free-energy change of –4.65 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?
Im not even sure where to start. Can you walk me through this please?
1 answer
1 answer