Use bond enthalpies in Table 5.4 to estimate \(\Delta H\) for each of the following reactions: (a) $\mathrm{H}-\mathrm{H}(g)+\mathrm{Br}-\mathrm{Br}(g) \longrightarrow 2 \mathrm{H}-\mathrm{Br}(g)$ (b)

Short Answer

Expert verified
To estimate the change in enthalpy, ΔH, for reaction (a), first identify the bonds formed and broken during the reaction: - Breaking: 1 H-H bond and 1 Br-Br bond - Forming: 2 H-Br bonds Next, use the bond enthalpy values from Table 5.4: - H-H bond: 436 kJ/mol - Br-Br bond: 193 kJ/mol - H-Br bond: 364 kJ/mol The equation for ΔH is: ΔH = (Bond enthalpies of bonds broken) - (Bond enthalpies of bonds formed) ΔH = [(1 × 436) + (1 × 193)] - (2 × 364) = 629 - 728 = -99 kJ/mol So, for reaction (a), ΔH = -99 kJ/mol.

Step by step solution

01

Identify the bonds formed and broken in the first reaction

In reaction (a), we have: - Breaking: 1 H-H bond and 1 Br-Br bond - Forming: 2 H-Br bonds
02

Write the equation for ΔH in terms of bond enthalpies

Using the bond enthalpies, we can write the equation for ΔH as: ΔH = (Bond enthalpies of bonds broken) - (Bond enthalpies of bonds formed)
03

Find the bond enthalpy values from Table 5.4

Look up the bond enthalpy values for the bonds involved in the reaction from Table 5.4: - H-H bond: 436 kJ/mol - Br-Br bond: 193 kJ/mol - H-Br bond: 364 kJ/mol
04

Calculate ΔH for the first reaction

Use the bond enthalpy values found in Table 5.4 to substitute into the ΔH equation: ΔH = [(1 × 436) + (1 × 193)] - (2 × 364) = 629 - 728 = -99 kJ/mol For reaction (a), ΔH = -99 kJ/mol
05

Identify the bonds formed and broken in the second reaction

In reaction (b), repeat steps 1 through 4, taking care to use the bond information given for the specific reaction.

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