Question

Using the table of average bond dissociation enthalpies at $25^{\circ} \mathrm{C}$, determine which of the following reactions are energetically favorable at room temperature. Assume that \(\Delta S=0\).

Short answer

Answer: Reactions with a negative change in enthalpy (ΔH < 0) are energetically favorable at room temperature. Calculate the change in enthalpy for each given reaction using the steps in the solution to determine if it is energetically favorable.

Step-by-step solution

Reaction 1:

\(A_1B_1 + A_2B_2 \rightarrow A_1A_2 + B_1B_2\)

Calculate energy required to break initial bonds.

Calculate the bond dissociation enthalpies for \(A_1B_1\) and \(A_2B_2\) and add them together to get the total energy required to break the bonds. Suppose the values are \(E_{1}\) and \(E_{2}\), then the energy required is \(E_{1} + E_{2}\).

Calculate energy released from forming new bonds.

Calculate the bond dissociation enthalpies for \(A_1A_2\) and \(B_1B_2\) and add them together to get the total energy released when the new bonds form. Suppose the values are \(E_{3}\) and \(E_{4}\), then the energy released is \(E_{3} + E_{4}\).

Calculate the change in enthalpy for Reaction 1.

Subtract the energy released from forming new bonds from the energy required to break initial bonds. If the result is negative, the reaction is energetically favorable. So, if \(E_{3} + E_{4} - (E_{1} + E_{2}) < 0\), Reaction 1 is energetically favorable at room temperature. Perform these steps for other reactions as well.

Reaction 2:

\(A_3B_3 + A_4B_4 \rightarrow A_3A_4 + B_3B_4\)

Calculate energy required to break initial bonds.

Calculate the bond dissociation enthalpies for \(A_3B_3\) and \(A_4B_4\) and add them together to get the total energy required to break the bonds. Suppose the values are \(E_{5}\) and \(E_{6}\), then the energy required is \(E_{5} + E_{6}\).

Calculate energy released from forming new bonds.

Calculate the bond dissociation enthalpies for \(A_3A_4\) and \(B_3B_4\) and add them together to get the total energy released when the new bonds form. Suppose the values are \(E_{7}\) and \(E_{8}\), then the energy released is \(E_{7} + E_{8}\).

Calculate the change in enthalpy for Reaction 2.

Subtract the energy released from forming new bonds from the energy required to break initial bonds. If the result is negative, the reaction is energetically favorable. So, if \(E_{7} + E_{8} - (E_{5} + E_{6}) < 0\), Reaction 2 is energetically favorable at room temperature. Repeat these steps for any other given reactions. After analyzing all reactions and calculating their changes in enthalpy, we can determine which reactions are energetically favorable at room temperature by checking if their change in enthalpy is negative.