Questions: Bromination of isobutane is a two-step reaction. Step 1 + Br . → + H-Br Step 2 Br-Br → + Br . Overall reaction Using the table of bond dissociation energies for A-B → A+B, calculate the enthalpy of each step and the enthalpy of the overall reaction. Bond broken ΔH, kJ / mol H-H 436 Br-Br 194 H-Br 366 (CH3)3C-H 400. (CH3)3C-Br 292

Bromination of isobutane is a two-step reaction. Step 1 + Br . → + H-Br Step 2 Br-Br → + Br . Overall reaction Using the table of bond dissociation energies for A-B → A+B, calculate the enthalpy of each step and the enthalpy of the overall reaction. Bond broken ΔH, kJ / mol H-H 436 Br-Br 194 H-Br 366 (CH3)3C-H 400. (CH3)3C-Br 292
Transcript text: Bromination of isobutane is a two-step reaction. Step 1 $+$ Br . $\rightarrow$ $+$ $\mathrm{H}-\mathrm{Br}$ Step 2 $\mathrm{Br}-\mathrm{Br}$ $\longrightarrow$ $+$ Br . Overall reaction Using the table of bond dissociation energies for $\mathrm{A}-\mathrm{B} \rightarrow \mathrm{A}+\mathrm{B}$, calculate the enthalpy of each step and the enthalpy of the overall reaction. \begin{tabular}{|c|c|} \hline Bond broken & $\Delta H, \mathbf{k J} / \mathrm{mol}$ \\ \hline $\mathrm{H}-\mathrm{H}$ & 436 \\ \hline $\mathrm{Br}-\mathrm{Br}$ & 194 \\ \hline $\mathrm{H}-\mathrm{Br}$ & 366 \\ \hline$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{H}$ & 400. \\ \hline$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}$ & 292 \\ \hline \end{tabular}
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Solution

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Solution Steps

Step 1: Calculate the enthalpy change for Step 1

The enthalpy change (ΔH) is calculated as the sum of bond energies broken minus the sum of bond energies formed. In Step 1, a (CH3)3C-H bond is broken and an H-Br bond is formed. Thus, ΔH1 = 400 kJ/mol - 366 kJ/mol = +34 kJ/mol.

Step 2: Calculate the enthalpy change for Step 2

In Step 2, a Br-Br bond is broken and a (CH3)3C-Br bond is formed. Thus, ΔH2 = 194 kJ/mol - 292 kJ/mol = -98 kJ/mol.

Step 3: Calculate the enthalpy change for the overall reaction

The overall enthalpy change is the sum of the enthalpy changes for each step: ΔHoverall = ΔH1 + ΔH2 = 34 kJ/mol + (-98 kJ/mol) = -64 kJ/mol.

Final Answer:

ΔH1 = +34 kJ/mol ΔH2 = -98 kJ/mol ΔHoverall = -64 kJ/mol

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