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Chapter 6: Problem 74

You are given the following data: $$ \begin{array}{c} \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g) \quad \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{Br}(g) \quad \Delta H^{\circ}=192.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g) \\ \Delta H^{\circ}=-72.4 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{H}(g)+\operatorname{Br}(g) \longrightarrow \operatorname{HBr}(g) $$

Short Answer

Expert verified
The enthalpy change (\( \Delta H^{\circ} \)) for the reaction \( \text{H}(g) + \text{Br}(g) \rightarrow \text{HBr}(g) \) is 428.4 kJ/mol.

Step by step solution

01

Understand the given information

The given reactions are: \n1. \( \text{H}_{2}(g) \rightarrow 2 \text{H}(g) \) with \( \Delta H^{\circ}=436.4 \text{~kJ/mol} \)\n2. \( \text{Br}_{2}(g) \rightarrow 2 \text{Br}(g) \) with \( \Delta H^{\circ}=192.5 \text{~kJ/mol} \)\n3. \( \text{H}_{2}(g)+\text{Br}_{2}(g) \rightarrow 2 \text{HBr}(g) \) with \( \Delta H^{\circ}=-72.4 \text{~kJ/mol} \)\nThe required reaction is \( \text{H}(g) + \text{Br}(g) \rightarrow \text{HBr}(g) \)
02

Use Hess's Law to write the required reaction as the sum of the given reactions

To form \( \text{H}(g) + \text{Br}(g) \rightarrow \text{HBr}(g) \), half of the first and second reactions will be added and the third reaction will be subtracted: \(\frac{1}{2} \times\)(first reaction) + \(\frac{1}{2} \times\)(second reaction) - (third reaction).
03

Calculate \( \Delta H^{\circ} \) for the required reaction

Apply the steps from Step 2 to find \( \Delta H^{\circ} \) for the required reaction:\(\Delta H^{\circ} = \frac{1}{2} \times\) (first reaction's \( \Delta H^{\circ} \)) + \(\frac{1}{2} \times\)(second reaction's \( \Delta H^{\circ} \)) - (third reaction's \( \Delta H^{\circ} \)) \(\Delta H^{\circ} = \frac{1}{2} \times 436.4 \text{~kJ/mol} + \frac{1}{2} \times 192.5\text{~kJ/mol} - (-72.4 \text{~kJ/mol})\)

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Most popular questions from this chapter

State Hess's law. Explain, with one example, the usefulness of Hess's law in thermochemistry.

Consider two metals A and B, each having a mass of \(100 \mathrm{~g}\) and an initial temperature of \(20^{\circ} \mathrm{C}\). The specific heat of \(A\) is larger than that of \(B\). Under the same heating conditions, which metal would take longer to reach a temperature of \(21^{\circ} \mathrm{C} ?\)

Consider the reaction $$ \begin{aligned} \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow & 2 \mathrm{NH}_{3}(g) \\\ \Delta H_{\mathrm{rxn}}^{\circ} &=-92.6 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ If 2.0 moles of \(\mathrm{N}_{2}\) react with 6.0 moles of \(\mathrm{H}_{2}\) to form \(\mathrm{NH}_{3},\) calculate the work done (in joules) against a pressure of 1.0 atm at \(25^{\circ} \mathrm{C}\). What is \(\Delta E\) for this reaction? Assume the reaction goes to completion.

I he enthaipy of combustion benzo1c acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be \(-3226.7 \mathrm{~kJ} / \mathrm{mol}\). When \(1.9862 \mathrm{~g}\) of benzoic acid are burned in a calorimeter, the temperature rises from \(21.84^{\circ} \mathrm{C}\) to \(25.67^{\circ} \mathrm{C}\). What is the heat capacity of the bomb? (Assume that the quantity of water surrounding the bomb is exactly \(2000 \mathrm{~g} .)\)

A 2.10 -mole sample of crystalline acetic acid, initially at \(17.0^{\circ} \mathrm{C}\), is allowed to melt at \(17.0^{\circ} \mathrm{C}\) and is then heated to \(118.1^{\circ} \mathrm{C}\) (its normal boiling point) at 1.00 atm. The sample is allowed to vaporize at \(118.1^{\circ} \mathrm{C}\) and is then rapidly quenched to \(17.0^{\circ} \mathrm{C}\), so that it recrystallizes. Calculate \(\Delta H^{\circ}\) for the total process as described.
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