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Chapter 7: Problem 15

How much heat, in kilojoules, is evolved in the complete combustion of (a) \(1.325 \mathrm{g} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm} ;\) (b) \(28.4 \mathrm{L} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(\mathrm{STP} ;(\mathrm{c})\) \(12.6 \mathrm{LC}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(23.6^{\circ} \mathrm{C}\) and \(738 \mathrm{mmHg} ?\) Assume that the enthalpy change for the reaction does not change significantly with temperature or pressure. The complete combustion of butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) is represented by the equation $$\begin{array}{r} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+5 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ}=-2877 \mathrm{kJ} \end{array}$$

Short Answer

Expert verified
The heat evolved for the combustion of \(1.325g \) of \(C_{4}H_{10}\) is \(-65.1kJ\), for \(28.4L\) of \(C_{4}H_{10}\) at \(STP\) is \(-3407.3kJ\), and for \(12.6L\) of \(C_{4}H_{10}\) at \(23.6^{\circ}C\) and \(738mmHg\) is \(-1534.6kJ\).

Step by step solution

01

Mole Calculation

Calculate the number of moles of \(C_{4}H_{10}\) using the formula: \(n=\frac{m}{MW}\), where \(m=1.325g\) is the mass and \(MW=58.122g/mol\) is the molecular weight of butane.
02

Heat Calculation (Part 1)

Calculate the heat evolved in the combustion using the formula: \( q = n \cdot ΔH^{\circ}\), where \(n\) is the number of moles calculated and \(-2877 kJ/mol\) is enthalpy change. The sign of the heat evolved is negative because it's an exothermic reaction.
03

Mole Calculation (Part 2)

For the second condition, calculate the number of moles from the volume using the formula: \(n=\frac{V}{V_{m}}\), where \(V=28.4L\) is the volume and \(V_{m}=22.414L/mol\) at standard conditions.
04

Heat Calculation (Part 2)

Repeat the heat calculation with the new moles calculated.
05

Mole Calculation (Part 3)

For the third condition, calculate volume at standard conditions using the Ideal Gas Law. Then, calculate the mole from the volume at the standard condition using the same formula from Step 3.
06

Heat Calculation (Part 3)

Again, repeat the heat computation with the calculated moles.

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Most popular questions from this chapter

A handbook lists two different values for the heat of combustion of hydrogen: \(33.88 \mathrm{kcal} / \mathrm{g} \mathrm{H}_{2}\) if \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is formed, and \(28.67 \mathrm{kcal} / \mathrm{g} \mathrm{H}_{2}\) if \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is formed. Explain why these two values are different, and indicate what property this difference represents. Devise a means of verifying your conclusions.

In a student experiment to confirm Hess's law, the reaction $$\mathrm{NH}_{3}(\text { concd aq })+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})$$ was carried out in two different ways. First, \(8.00 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\text { aq })\) was added to \(100.0 \mathrm{mL}\) of 1.00 M HCl in a calorimeter. [The NH \(_{3}(\) aq) was slightly in excess.] The reactants were initially at \(23.8^{\circ} \mathrm{C},\) and the final temperature after neutralization was \(35.8^{\circ} \mathrm{C} .\) In the second experiment, air was bubbled through \(100.0 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\mathrm{aq})\) sweeping out \(\mathrm{NH}_{3}(\mathrm{g})\) (see sketch). The \(\mathrm{NH}_{3}(\mathrm{g})\) was neutralized in \(100.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). The temperature of the concentrated \(\mathrm{NH}_{3}(\text { aq })\) fell from 19.3 to \(13.2^{\circ} \mathrm{C} .\) At the same time, the temperature of the 1.00 M HCl rose from 23.8 to 42.9 ^ C as it was neutralized by \(\mathrm{NH}_{3}(\mathrm{g}) .\) Assume that all solutions have densities of \(1.00 \mathrm{g} / \mathrm{mL}\) and specific heats of \(4.18 \mathrm{Jg}^{-1 \circ} \mathrm{C}^{-1}\) (a) Write the two equations and \(\Delta H\) values for the processes occurring in the second experiment. Show that the sum of these two equations is the same as the equation for the reaction in the first experiment. (b) Show that, within the limits of experimental error, \(\Delta H\) for the overall reaction is the same in the two experiments, thereby confirming Hess's law.

A clay pot containing water at \(25^{\circ} \mathrm{C}\) is placed in the shade on a day in which the temperature is \(30^{\circ} \mathrm{C} .\) The outside of the clay pot is kept moist. Will the temperature of the water inside the clay pot (a) increase; (b) decrease; (c) remain the same?

Determine \(\Delta H^{\circ}\) for this reaction from the data below. \(\mathrm{N}_{2} \mathrm{H}_{4}(1)+2 \mathrm{H}_{2} \mathrm{O}_{2}(1) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(1)\) $$\begin{array}{r} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-622.2 \mathrm{kJ} \end{array}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(1) \quad \Delta H^{\circ}=-187.8 \mathrm{kJ}$$

Refer to Example \(7-3 .\) Based on the heat of combustion of sucrose established in the example, what should be the temperature change \((\Delta T)\) produced by the combustion of \(1.227 \mathrm{g} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) in a bomb calorimeter assembly with a heat capacity of \(3.87 \mathrm{kJ} /^{\circ} \mathrm{C} ?\)
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