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Chapter 22: Problem 158

Estimate \(\Delta H\) for the following reactions using bond energies given in Table \(8.5 .\) $$ 3 \mathrm{CH}_{2}=\mathrm{CH}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 3 \mathrm{CH}_{3}-\mathrm{CH}_{3}(g) $$ The enthalpies of formation for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) and \(\mathrm{C}_{6} \mathrm{H}_{12}(g)\) are 82.9 and $-90.3 \mathrm{kJ} / \mathrm{mol}\( , respectively. Calculate \)\Delta H^{\circ}$ for the two reactions using standard enthalpies of formation from Appendix \( 4 .\) Account for any differences between the results obtained from the two methods.

Short Answer

Expert verified
In the given reaction, the enthalpy change using bond energies is calculated as: \[ \Delta H = 3 \times \text{C=C} + 6 \times \text{H-H} - 6 \times \text{C-C} - 12 \times \text{C-H} \] Using standard enthalpies of formation, we obtained: \[ \Delta H^{\circ} = 3 \times \mathrm{C}_{6} \mathrm{H}_{12}(g) - 3 \times \mathrm{C}_{6} \mathrm{H}_{6}(g) \] The difference between these two results can be attributed to the use of average bond energies in the bond energy method, while specific bond energies are considered in the standard enthalpies of formation method.

Step by step solution

01

Identify the bonds broken and formed

In the given reaction, the following bonds are broken: - 3 C=C in CH2=CH2 - 6 H-H in 3 H2 molecules And the following bonds are formed: - 6 C-C in 3 CH3-CH3 molecules - 12 C-H in 3 CH3-CH3 molecules
02

Use Table 8.5 to find the bond energies

Look up the bond energies in Table 8.5 for the following bonds: - C=C - H-H - C-C - C-H
03

Calculate \(\Delta H\) using bond energies

Apply the following formula to calculate \(\Delta H\): \[ \Delta H = \Sigma \text{(Bond Energies of reactants)} - \Sigma \text{(Bond Energies of products)} \] Substitute the values from Step 2: \[ \Delta H = 3 \times \text{C=C} + 6 \times \text{H-H} - 6 \times \text{C-C} - 12 \times \text{C-H} \]
04

Calculate \(\Delta H^{\circ}\) for the reactions using standard enthalpies of formation

Given the enthalpies of formation for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) and \(\mathrm{C}_{6} \mathrm{H}_{12}(g)\). Use the following formula to calculate \(\Delta H^{\circ}\): \[ \Delta H^{\circ} = \Sigma \text{(Standard enthalpies of formation for products)} - \Sigma \text{(Standard enthalpies of formation for reactants)} \] Substitute the values: \[ \Delta H^{\circ} = 3 \times \mathrm{C}_{6} \mathrm{H}_{12}(g) - (3 \times \mathrm{C}_{6} \mathrm{H}_{6}(g) + 3 \times \mathrm{H}_{2}(g)) \] (Note: The standard enthalpy of formation for any element in its most stable state, like \(\mathrm{H}_{2}(g)\), is zero.)
05

Compare the results from both methods

Compare the \(\Delta H\) calculated from bond energies in Step 3 with the \(\Delta H^{\circ}\) calculated from standard enthalpies of formation in Step 4. Discuss their differences and explain the possible reasons for the difference (e.g., the use of average bond energies in the first method compared to specific bond energies in the second method).

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