/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 114 The sugar arabinose, \(\mathrm{C... [FREE SOLUTION] | 91Ó°ÊÓ

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The sugar arabinose, \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5},\) is burned completely in oxygen in a calorimeter. $$ \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}(s)+5 \mathrm{O}_{2}(g) \longrightarrow 5 \mathrm{CO}_{2}(g)+5 \mathrm{H}_{2} \mathrm{O}(l) $$ Burning a 0.548 -g sample caused the temperature to rise from \(20.00^{\circ} \mathrm{C}\) to \(20.54^{\circ} \mathrm{C}\). The heat capacity of the calorimeter and its contents is \(15.8 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). Calculate \(\Delta H\) for the combustion reaction per mole of arabinose.

Short Answer

Expert verified
\( \Delta H = -2338.36 \text{ kJ/mol} \) for the combustion of arabinose.

Step by step solution

01

Calculating Heat Absorbed by Calorimeter

Use the formula \( q = C \times \Delta T \), where \( C = 15.8 \text{ kJ/}^\circ\text{C} \) and \( \Delta T = 20.54^\circ\text{C} - 20.00^\circ\text{C} = 0.54^\circ\text{C} \). Therefore, the heat absorbed, \( q \), is \( 15.8 \text{ kJ/}^\circ\text{C} \times 0.54^\circ\text{C} = 8.532 \text{ kJ} \).
02

Calculating Moles of Arabinose

Find the molar mass of arabinose (\( \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5} \)). The atomic masses are: C = 12.01, H = 1.01, O = 16.00. Thus, molar mass = \( 5(12.01) + 10(1.01) + 5(16.00) = 150.15 \text{ g/mol} \). Calculate moles using the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \) = \( \frac{0.548}{150.15} = 0.00365 \text{ moles} \).
03

Calculating ΔH for the Reaction per Mole

The heat change is \( -q \) as it is an exothermic reaction: \( \Delta H = \frac{-q}{\text{moles of arabinose}} = \frac{-8.532 \text{ kJ}}{0.00365 \text{ moles}} \). Compute this to find \( \Delta H = -2338.36 \text{ kJ/mol} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
In chemical reactions, the enthalpy change (\( \Delta H \)) represents the heat absorbed or released under constant pressure. It's a key concept in calorimetry, as it helps us understand the energy dynamics of reactions like combustion. For exothermic reactions, such as burning arabinose, the system releases heat, and \( \Delta H \) becomes negative. This reflects heat flow out of the system and into the surroundings.
To calculate the enthalpy change during this combustion, we need to determine how much heat the reaction generated. This is done by measuring the temperature change in a calorimeter and using it to find the heat released. By dividing the heat released by the number of moles of arabinose burned, we can find \( \Delta H \) in terms of energy per mole. This provides a standardized measure of the energy change for this particular reaction under the given conditions.
Combustion Reaction
A combustion reaction is a chemical process where a substance combines with oxygen, releasing energy in the form of heat and light. In the case of arabinose combustion, we have:
  • Reactants: Arabinose \( (\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}) \) and oxygen \( (\mathrm{O}_{2}) \)
  • Products: Carbon dioxide \( (\mathrm{CO}_{2}) \) and water \( (\mathrm{H}_{2}\mathrm{O}) \)
The balanced equation ensures that all oxygen atoms are accounted for as they convert arabinose completely into carbon dioxide and water.
Combustion reactions are exothermic, meaning they release heat. This is essential for understanding how the calorimeter absorbs the heat produced, subsequently allowing us to measure the heat of reaction. Recognizing the nature of the reaction helps explain why the calorimeter registered a temperature increase.
Calorimeter Calculation
To measure the energy released during a combustion reaction, a calorimeter is used. This device traps and measures the heat released, allowing for precise calculation of thermal changes. The specific formula used here is \( q = C \times \Delta T \), where:
  • \( q \) is the heat absorbed by the calorimeter.
  • \( C \) is the heat capacity of the calorimeter, representing its ability to absorb heat.
  • \( \Delta T \) is the change in temperature.
In our example, using the calorimeter's heat capacity and temperature change, we compute the heat absorbed. This heat value is critical for determining the reaction's energy profile. It shows how much energy the reaction produces, as indicated by the temperature rise from \( 20.00^{\circ} \mathrm{C} \) to \( 20.54^{\circ} \mathrm{C} \). This change confirms the energy release from the arabinose combustion.
Molar Mass of Arabinose
Knowing the molar mass of arabinose is essential for understanding how much of this substance participates in the reaction. The molecular formula \( \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5} \) specifies its composition, comprising carbon, hydrogen, and oxygen.
Calculating the molar mass requires summing up the atomic masses of all atoms in the molecule:
  • Carbon: \( 5 \times 12.01 \text{ g/mol} \)
  • Hydrogen: \( 10 \times 1.01 \text{ g/mol} \)
  • Oxygen: \( 5 \times 16.00 \text{ g/mol} \)
This adds up to a total molar mass of \( 150.15 \text{ g/mol} \).
Understanding this helps convert the mass of the substance involved in the reaction to moles, a critical step in calculating the enthalpy change per mole. Thus, knowing the molar mass helps relate the physical amount of the reactant to the energy change of the chemical process.

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Most popular questions from this chapter

Hydrogen, \(\mathrm{H}_{2}\), is used as a rocket fuel. The hydrogen is burned in oxygen to produce water vapor. $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H=-484 \mathrm{~kJ} $$ What is the enthalpy change per kilogram of hydrogen?

Define the heat capacity of a substance. Define the specific heat of a substance.

When white phosphorus burns in air, it produces phosphorus(V) oxide. $$ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) ; \Delta H=-3010 \mathrm{~kJ} $$ What is \(\Delta H\) for the following equation? $$ \mathrm{P}_{4} \mathrm{O}_{10}(s) \longrightarrow \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) $$

When ice at \(0^{\circ} \mathrm{C}\) melts to liquid water at \(0^{\circ} \mathrm{C}\), it absorbs \(0.334 \mathrm{~kJ}\) of heat per gram. Suppose the heat needed to melt \(38.0 \mathrm{~g}\) of ice is absorbed from the water contained in a glass. If this water has a mass of \(0.210 \mathrm{~kg}\) and a temperature of \(21.0^{\circ} \mathrm{C}\), what is the final temperature of the water? (Note that you will also have \(38.0 \mathrm{~g}\) of water at \(0^{\circ} \mathrm{C}\) from the ice.)

A \(29.1-\mathrm{mL}\) sample of \(1.05 \mathrm{M} \mathrm{KOH}\) is mixed with \(20.9 \mathrm{~mL}\) of \(1.07 M \mathrm{HBr}\) in a coffee-cup calorimeter (see Section 6.6 of your text for a description of a coffee-cup calorimeter). The enthalpy of the reaction, written with the lowest whole-number coefficients, is \(-55.8 \mathrm{~kJ}\). Both solutions are at \(21.8^{\circ} \mathrm{C}\) prior to mixing and reacting. What is the final temperature of the reaction mixture? When solving this problem, assume that no heat is lost from the calorimeter to the surroundings, the density of all solutions is \(1.00 \mathrm{~g} / \mathrm{mL},\) and volumes are additive.

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