Certainly! Let's go through these calculations step-by-step.
For step 1: The reaction involves the following bonds being broken and formed:
[tex]\[ \text{(CH}_3)_3\text{C-H} + \text{Br-Br} \rightarrow \text{(CH}_3)_3\text{C.} + \text{Br.} \][/tex]
Bonds Broken:
1. One [tex]\((\text{CH}_3)_3\text{C-H}\)[/tex] bond
2. One [tex]\(\text{Br-Br}\)[/tex] bond
The bond dissociation energies for these bonds from the table are:
[tex]\[
\begin{align*}
(\text{CH}_3)_3\text{C-H} & : 400 \, \text{kJ/mol} \\
\text{Br-Br} & : 194 \, \text{kJ/mol}
\end{align*}
\][/tex]
Adding these two values gives the total bond energy broken in step 1:
[tex]\[
\text{Energy of bonds broken in Step 1} = 400 \, \text{kJ/mol} + 194 \, \text{kJ/mol} = 594 \, \text{kJ/mol}
\][/tex]
Bonds Formed:
1. One [tex]\((\text{CH}_3)_3\text{C-Br}\)[/tex] bond
2. One [tex]\(\text{H-Br}\)[/tex] bond
The bond dissociation energies for these bonds from the table are:
[tex]\[
\begin{align*}
(\text{CH}_3)_3\text{C-Br} & : 292 \, \text{kJ/mol} \\
\text{H-Br} & : 366 \, \text{kJ/mol}
\end{align*}
\][/tex]
Adding these two values gives the total bond energy formed in step 1:
[tex]\[
\text{Energy of bonds formed in Step 1} = 292 \, \text{kJ/mol} + 366 \, \text{kJ/mol} = 658 \, \text{kJ/mol}
\][/tex]
Enthalpy change (ΔH) for Step 1:
The enthalpy change for the reaction is calculated by subtracting the energy of the bonds formed from the energy of the bonds broken:
[tex]\[
\Delta H_{\text{Step 1}} = \text{Energy of bonds broken} - \text{Energy of bonds formed}
\][/tex]
Substitute the values:
[tex]\[
\Delta H_{\text{Step 1}} = 594 \, \text{kJ/mol} - 658 \, \text{kJ/mol} = -64 \, \text{kJ/mol}
\][/tex]
Thus, the enthalpy change for step 1 ([tex]\(\Delta H_{\text{Step 1}}\)[/tex]) is [tex]\(-64 \, \text{kJ/mol}\)[/tex].
Note: Since the question does not provide details about Step 2, we focus on Step 1 only. To determine the enthalpy of the overall reaction, details of Step 2 would be needed.
In summary:
[tex]\[
\Delta H_{\text{Step 1}} = -64 \, \text{kJ/mol}
\][/tex]
