Question

1. Consider the reaction A(aq) ⇆ 2B(aq), and assume that you dissolve a given amount of B in pure water. Explain whether these statements are true or false. If false, write the correct statement.

   • The units of the equilibrium constant are M (molar).

   • [B]2 /(1M.[A]) = Keq at all times

   • [B]2 /(1M.[A]) = Keq only when equilibrium is reached

2. Consider the reaction of dissociation of duplex DNA. What are the signs of ∆H and ∆S? Provide a justification. Will this reaction be favored at low or high temperatures?

3. Consider a reaction with ∆H > 0 and ∆S > 0. Will this reaction be favored at low or high temperatures?

Answer 1

Part 1. [B]2/(1M[A]) = Keq only when equilibrium is reached. This statement is true and the other statement is false.

Explanation: Since B is added to water, it will start reacting to form A. Only when sufficient A has been formed, equilibrium will be reached and the ratio of their concentration will be constant. Therefore the ratio cannot be constant at all times. Hence first statement is not correct.

Part 2. For this process, Delta H will be positive and Delta S will be negative.

The reaction will be favored at high temperature.

Explanation: Double strand DNA is stable due to hydrogen bonding and pi stacking of aromatic rings. When they are separated, energy will increase resulting in positive enthalpy change. When single strand DNA is formed, it will have more flexibility and thus more conformational entropy. Hence entropy change will be positive. To make Delta G negative, the entropic contribution has to be high. Therefore if temperature is higher, then in the relation

G = H - TS

The TS term will be higher and more negative and hence higher temperature will be more favorable.

Part 3. This reaction will be favored at high temperature. The explanation is same as the above as there also Delta H > 0 and Delta S > 0.