A reaction has a standard free-energy change of -4.36 kJ/mol at 25 \"C. What are

Question

A reaction has a standard free-energy change of -4.36 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? How would your answers above change if the reaction had a standard free-Energy change of +4.36 kJ/mol? There would be less A and 8 but more C. All concentrations would be higher. There would be no change to the answers. There would be more A and 8 but less C. All concentrations would be lower.

Explanation / Answer

standard free energy = - RTlnK lnK = 4360/8.31 * 298 = 1.76 K = 5.81 now , at equilibrium we have, 5.81 = 2x/(0.3-x)*(0.4-x) x = 0.13 hence equilibrium concentration of A = (0.3 - 0.13) = 0.17 M equilibrium concentration of B = (0.4 - 0.13) = 0.27 M equilibrium concentration of C = 2*0.13 = 0.26 M If +4.36 kJ/mol then, there would be more A and B and less C, since +ve free energy does not favour forward reaction.

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