Solve for [tex]\( x \)[/tex].
[tex]\[ 3x = 6x - 2 \][/tex]




Calculate the enthalpy of each step and the enthalpy of the overall reaction for the bromination of isobutane, a two-step reaction.

Using the table of bond dissociation energies for [tex]\( A - B \longrightarrow A + B \)[/tex]:
[tex]\[
\begin{tabular}{|c|c|}
\hline
\text{Bond broken} & \(\Delta H \, (kJ/mol)\) \\
\hline
H - H & 436 \\
\hline
Br - Br & 194 \\
\hline
H - Br & 366 \\
\hline
(CH_3)_3C - H & 400 \\
\hline
(CH_3)_3C - Br & 292 \\
\hline
\end{tabular}
\][/tex]

Step 1: [tex]\( \Delta H = \square \)[/tex]



Answer :

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]