Here are the solutions for the enthalpy worksheet problems. I have broken down each reaction into bonds broken and bonds made, calculated the total energy for each, and found the final enthalpy change ($\Delta H$).
1. Reaction: $\text{H}_2 + \text{Br}_2 \rightarrow 2\text{HBr}$
Step 1: Identify Bonds Broken (Reactants)
*
$\text{H}-\text{H}$: 1 bond $\times$ $436 \text{ kJ/mol} = 436 \text{ kJ}$
*
$\text{Br}-\text{Br}$: 1 bond $\times$ $193 \text{ kJ/mol} = 193 \text{ kJ}$
*
Total Energy In: $436 + 193 = \mathbf{629 \text{ kJ}}$
Step 2: Identify Bonds Made (Products)
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$\text{H}-\text{Br}$: 2 bonds $\times$ $366 \text{ kJ/mol} = 732 \text{ kJ}$
*
Total Energy Out: $\mathbf{732 \text{ kJ}}$
Step 3: Calculate Enthalpy Change
* $\Delta H = \text{Energy In} - \text{Energy Out}$
* $\Delta H = 629 - 732 = \mathbf{-103 \text{ kJ/mol}}$
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2. Reaction: $\text{H}_2 + \text{F}_2 \rightarrow 2\text{HF}$
Step 1: Identify Bonds Broken (Reactants)
*
$\text{H}-\text{H}$: 1 bond $\times$ $436 \text{ kJ/mol} = 436 \text{ kJ}$
*
$\text{F}-\text{F}$: 1 bond $\times$ $158 \text{ kJ/mol} = 158 \text{ kJ}$
*
Total Energy In: $436 + 158 = \mathbf{594 \text{ kJ}}$
Step 2: Identify Bonds Made (Products)
*
$\text{H}-\text{F}$: 2 bonds $\times$ $568 \text{ kJ/mol} = 1136 \text{ kJ}$
*
Total Energy Out: $\mathbf{1136 \text{ kJ}}$
Step 3: Calculate Enthalpy Change
* $\Delta H = \text{Energy In} - \text{Energy Out}$
* $\Delta H = 594 - 1136 = \mathbf{-542 \text{ kJ/mol}}$
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3. Reaction: $\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$
Step 1: Identify Bonds Broken (Reactants)
*
$\text{C}-\text{H}$: Propane ($\text{C}_3\text{H}_8$) has 8 $\text{C}-\text{H}$ bonds.
* $8 \times 413 \text{ kJ/mol} = 3304 \text{ kJ}$
*
$\text{C}-\text{C}$: Propane has 2 $\text{C}-\text{C}$ bonds.
* $2 \times 348 \text{ kJ/mol} = 696 \text{ kJ}$
*
$\text{O}=\text{O}$: Oxygen ($5\text{O}_2$) has 5 double bonds.
* $5 \times 495 \text{ kJ/mol} = 2475 \text{ kJ}$
*
Total Energy In: $3304 + 696 + 2475 = \mathbf{6475 \text{ kJ}}$
Step 2: Identify Bonds Made (Products)
*
$\text{C}=\text{O}$: Carbon dioxide ($3\text{CO}_2$) has 2 double bonds per molecule. Total = $3 \times 2 = 6$ bonds.
* $6 \times 799 \text{ kJ/mol} = 4794 \text{ kJ}$
*
$\text{O}-\text{H}$: Water ($4\text{H}_2\text{O}$) has 2 single bonds per molecule. Total = $4 \times 2 = 8$ bonds.
* $8 \times 463 \text{ kJ/mol} = 3704 \text{ kJ}$
*
Total Energy Out: $4794 + 3704 = \mathbf{8498 \text{ kJ}}$
Step 3: Calculate Enthalpy Change
* $\Delta H = \text{Energy In} - \text{Energy Out}$
* $\Delta H = 6475 - 8498 = \mathbf{-2023 \text{ kJ/mol}}$
Final Answer:
1.
-103 kJ/mol
2.
-542 kJ/mol
3.
-2023 kJ/mol
Parent Tip: Review the logic above to help your child master the concept of enthalpy worksheet.
