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Chapter 6: Problem 105

Combustion of table sugar produces \(\mathrm{CO}_{2}(g)\) and $\mathrm{H}_{2} \mathrm{O}(l) .$ When 1.46 \(\mathrm{g}\) table sugar is combusted in a constant-volume (bomb) calorimeter, 24.00 \(\mathrm{kJ}\) of heat is liberated. a. Assuming that table sugar is pure sucrose, $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)$ write the balanced equation for the combustion reaction. b. Calculate \(\Delta E\) in $\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$ for the combustion reaction of sucrose.

Short Answer

Expert verified
The balanced equation for the combustion of sucrose is C12H22O11(s) + 12O2(g) -> 12CO2(g) + 11H2O(l), and the change in internal energy per mole of sucrose combusted is 5,637.09 kJ/mol.

Step by step solution

01

Write the unbalanced combustion equation for sucrose

The combustion reaction involves the reaction of a substance with oxygen to produce carbon dioxide and water. The unbalanced chemical equation for the combustion of sucrose is: C12H22O11(s) + O2(g) -> CO2(g) + H2O(l)
02

Balance the combustion equation of sucrose

To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. We can do this by adjusting the coefficients in front of the compounds: C12H22O11(s) + 12O2(g) -> 12CO2(g) + 11H2O(l) Now the equation is balanced.
03

Determine the number of moles of sucrose combusted

We are given that 1.46 g of table sugar (sucrose) has been combusted. To determine the number of moles, we can use the molecular weight of sucrose, which is 342.3 g/mol: Moles of sucrose = (mass of sucrose) / (molecular weight of sucrose) Moles of sucrose = 1.46 g / 342.3 g/mol Moles of sucrose = 0.00426 mol
04

Calculate the change in internal energy per mole of sucrose combusted

We are given that 24.00 kJ of heat was liberated during the combustion of 1.46 g of sucrose. We also know that the number of moles of sucrose combusted is 0.00426 mol. ΔE = (heat liberated during combustion) / (number of moles of sucrose combusted) ΔE = 24.00 kJ / 0.00426 mol ΔE = 5637.09 kJ/mol Hence, the change in internal energy per mole of sucrose combusted is 5,637.09 kJ/mol.

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Most popular questions from this chapter

For the process $\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\( at 298 \)\mathrm{K}\( and \)1.0 \mathrm{atm},\( \)\Delta H$ is more positive than \(\Delta E\) by 2.5 \(\mathrm{kJ} / \mathrm{mol}\) . What does the 2.5 \(\mathrm{kJ} / \mathrm{mol}\) quantity represent?

The enthalpy of neutralization for the reaction of a strong acid with a strong base is \(-56 \mathrm{kJ} / \mathrm{mol}\) water produced. How much energy will be released when 200.0 \(\mathrm{mL}\) of 0.400 \(\mathrm{M} \mathrm{HNO}_{3}\) is mixed with 150.0 \(\mathrm{mL}\) of 0.500 \(\mathrm{M} \mathrm{KOH}\) ?

On Easter Sunday, April \(3,1983,\) nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: $$ 2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) $$ a. Calculate \(\Delta H^{\circ}\) for this reaction. Approximately \(2.0 \times\) \(10^{4}\) gal nitric acid was spilled. Assume that the acid was an aqueous solution containing 70.0\(\% \mathrm{HNO}_{3}\) by mass with a density of 1.42 \(\mathrm{g} / \mathrm{cm}^{3} .\) What mass of sodium car- bonate was required for complete neutralization of the spill, and what quantity of heat was evolved? ( \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{NaNO}_{3}(a q)=-467 \mathrm{kJ} / \mathrm{mol} )\) b. According to The Denver Post for April \(4,1983,\) authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of \(\Delta H^{\circ},\) what was their major concern?

The energy content of food is typically determined using a bomb calorimeter. Consider the combustion of a \(0.30-\mathrm{g}\) sample of butter in a bomb calorimeter having a heat capacity of 2.67 \(\mathrm{kJ}^{\prime} \mathrm{C}\) . If the temperature of the calorimeter increases from \(23.5^{\circ} \mathrm{C}\) to \(27.3^{\circ} \mathrm{C}\) , calculate the energy of combustion per gram of butter.

Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for $\mathrm{C}_{4} \mathrm{H}_{4}(-2341 \mathrm{kJ} / \mathrm{mol}), \mathrm{C}_{4} \mathrm{H}_{8}\( \)(-2755 \mathrm{kJ} / \mathrm{mol}),\( and \)\mathrm{H}_{2}(-286 \mathrm{kJ} / \mathrm{mol}),\( calculate \)\Delta H$ for the reaction $$ \mathrm{C}_{4} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g) $$
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