A reaction: A(aq) +B(aq) -><- C(aq) Has a standard free-energy change of -4.41 k

Question

A reaction:
A(aq) +B(aq) -><- C(aq)

Has a standard free-energy change of -4.41 kJ/mol at 25 degrees Celcius. 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?

[A]=? [B]=? [C]=?

How would the concentrations change if the reaction had a standard free-energy change of +4.41 kJ/mol?

A. There would be less A and B but more C.

B.All concentrations would be lower.

C. All concentrations would be higher.

D. There would be no change to the answers.

E. There would be more A and B but less C

Explanation / Answer

The reactants or products shift based on the number of moles on each side. For example, in this equation (assuming -- means -->) you have 2 moles on the left side, and 2 moles on the right side, so an equal amount of moles on each side. Therefore, an increase of A would result in a shift to the right (A is a reactant, and an increase in reactants will increase the products). Increase in B will shift right for the same reasons. An increase in C would shift left to maintain equilibrium. A decrease in A would shift left for the same reason, as would a decrease in B. A decrease in C would shift right (same reasoning--you want to maintain the same amount of moles in the right side, so if you decrease the right side, the reaction will 'fix itself' by shifting right). Doubling A and halving B should be no shift, and if you double B and C you would have 3 moles on the left (1 mol A + 2 mol B) and 4 moles on the right (4 moles C) so the reaction would shift left to make up for the higher amount of moles on the right side.

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