Question

h−c≡c−h(g) + 2hi(g) → (h₃c−c−i with h atoms) (g) using the table of average bond energies below, the δh for the reaction is ____ kj. options: -160, +160, +217, -217, -63

Explanation:

To calculate the enthalpy change ($\Delta H$) for a reaction using bond energies, we use the formula:
$\Delta H = \sum (\text{bond energies of bonds broken}) - \sum (\text{bond energies of bonds formed})$

Step 1: Identify bonds broken and formed

• Bonds broken:
• 1 $\ce{C\equiv C}$ triple bond (in $\ce{HC\equiv CH}$)
• 1 $\ce{C-H}$ bond (in $\ce{HC\equiv CH}$)
• 2 $\ce{H-I}$ bonds (in $2\ce{HI}$)

• Bonds formed:
• 1 $\ce{C-C}$ single bond (in the product)
• 2 $\ce{C-H}$ bonds (new, in the product)
• 2 $\ce{C-I}$ bonds (in the product)

Step 2: Recall average bond energies (standard values)

• $\ce{C\equiv C}$: $837\ \text{kJ/mol}$
• $\ce{C-H}$: $413\ \text{kJ/mol}$
• $\ce{H-I}$: $299\ \text{kJ/mol}$
• $\ce{C-C}$: $348\ \text{kJ/mol}$
• $\ce{C-I}$: $240\ \text{kJ/mol}$

Step 3: Calculate energy for bonds broken

Sum of bond energies for bonds broken:

$$\begin{align*} \text{Bonds broken} &= (\ce{C\equiv C}) + (\ce{C-H}) + 2(\ce{H-I}) \\ &= 837 + 413 + 2(299) \\ &= 837 + 413 + 598 \\ &= 1848\ \text{kJ/mol} \end{align*}$$

Step 4: Calculate energy for bonds formed

Sum of bond energies for bonds formed:

$$\begin{align*} \text{Bonds formed} &= (\ce{C-C}) + 2(\ce{C-H}) + 2(\ce{C-I}) \\ &= 348 + 2(413) + 2(240) \\ &= 348 + 826 + 480 \\ &= 1654\ \text{kJ/mol} \end{align*}$$

Step 5: Calculate $\Delta H$

Using $\Delta H = \text{bonds broken} - \text{bonds formed}$:

$$\begin{align*} \Delta H &= 1848 - 1654 \\ &= 194\ \text{kJ/mol} \quad \text{(Wait, this doesn’t match options. Did I miss a bond?)} \end{align*}$$

Wait, re-examining the reaction:
Reactant: $\ce{HC\equiv CH}$ (has 2 $\ce{C-H}$ bonds? Wait, $\ce{HC\equiv CH}$ is ethyne: $\ce{H-C\equiv C-H}$, so 2 $\ce{C-H}$ bonds! Oh, I made a mistake earlier. Let’s correct:

Step 1 (corrected): Bonds broken

• 1 $\ce{C\equiv C}$
• 2 $\ce{C-H}$ (in $\ce{HC\equiv CH}$)
• 2 $\ce{H-I}$

Step 3 (corrected): Bonds broken energy

$$\begin{align*} \text{Bonds broken} &= 837 + 2(413) + 2(299) \\ &= 837 + 826 + 598 \\ &= 2261\ \text{kJ/mol} \end{align*}$$

Step 4 (corrected): Bonds formed

Product: $\ce{I-CH2-CH2-I}$ (ethane derivative with 2 $\ce{C-I}$ and 4 $\ce{C-H}$? Wait, no: the product is $\ce{I-C-C-I}$ with four $\ce{C-H}$ bonds? Wait, the structure is $\ce{H2C-I - I-CH2}$? Wait, the given product is $\ce{H H | | I-C-C-I | | H H}$, so it’s $\ce{CH2I-CH2I}$? Wait, no: the structure shows $\ce{H H}$ on each $\ce{C}$, so $\ce{CH2I-CH2I}$? Wait, no, the structure is $\ce{H H | | I-C-C-I | | H H}$, so each $\ce{C}$ has two $\ce{H}$ and one $\ce{I}$? Wait, no: $\ce{C}$ has four bonds. So $\ce{C}$ is bonded to two $\ce{H}$, one $\ce{C}$, and one $\ce{I}$. So each $\ce{C}$ has: 2 $\ce{C-H}$, 1 $\ce{C-C}$, 1 $\ce{C-I}$. So for two $\ce{C}$ atoms:

• 1 $\ce{C-C}$
• 4 $\ce{C-H}$ (2 per $\ce{C}$)
• 2 $\ce{C-I}$ (1 per $\ce{C}$)

Step 4 (corrected): Bonds formed energy

$$\begin{align*} \text{Bonds formed} &= 348 + 4(413) + 2(240) \\ &= 348 + 1652 + 480 \\ &= 2480\ \text{kJ/mol} \end{align*}$$

Step 5 (corrected): $\Delta H$

$$\begin{align*} \Delta H &= 2261 - 2480 \\ &= -219\ \text{kJ/mol} \approx -217\ \text{kJ/mol} \quad \text{(matches the option -217)} \end{align*}$$

Answer:

-217