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Suppose you burn 0.300 g of \(C\) (graphite) in an excess of \(\mathrm{O}_{2}(\mathrm{g})\) in a constant volume calorimeter to give \(\mathrm{CO}_{2}(\mathrm{g})\). $$\mathrm{C}(\text { graphite })+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})$$ The temperature of the calorimeter, which contains \(775 \mathrm{g}\) of water, increases from \(25.00^{\circ} \mathrm{C}\) to \(27.38^{\circ} \mathrm{C} .\) The heat capacity of the bomb is \(893 \mathrm{J} / \mathrm{K}\). What quantity of heat is evolved per mole of carbon?

Short Answer

Expert verified
The heat evolved is approximately 393.9 kJ/mol of carbon.

Step by step solution

01

Calculate the Temperature Change

First, determine the change in temperature of the calorimeter. This is given by \[\Delta T = T_{\text{final}} - T_{\text{initial}} = 27.38^\circ C - 25.00^\circ C = 2.38^\circ C.\]
02

Calculate Heat Absorbed by Water

Next, calculate the heat absorbed by the water. Use the formula \[ q_{\text{water}} = m_{\text{water}} \times c \times \Delta T \]where:- \( m_{\text{water}} = 775 \text{ g} \) (mass of water),- \( c = 4.184 \text{ J/g}^\circ\text{C} \) (specific heat capacity of water),- \( \Delta T = 2.38^\circ C \).Substitute the values:\[q_{\text{water}} = 775 \text{ g} \times 4.184 \text{ J/g}^\circ\text{C} \times 2.38^\circ C = 7722.85 \text{ J}.\]
03

Calculate Heat Absorbed by Calorimeter

Calculate the heat absorbed by the calorimeter using the equation:\[ q_{\text{calorimeter}} = C_{\text{bomb}} \times \Delta T \]where- \( C_{\text{bomb}} = 893 \text{ J/K} \),- \( \Delta T = 2.38^\circ C \).Substitute the values:\[q_{\text{calorimeter}} = 893 \text{ J/K} \times 2.38 \text{ K} = 2124.34 \text{ J}.\]
04

Calculate Total Heat Evolved

The total heat evolved by the reaction is the sum of the heat absorbed by the water and the heat absorbed by the calorimeter:\[q_{\text{total}} = q_{\text{water}} + q_{\text{calorimeter}} = 7722.85 \text{ J} + 2124.34 \text{ J} = 9847.19 \text{ J}.\]
05

Convert to Heat Evolved per Mole of Carbon

Calculate the number of moles of carbon in \(0.300 \text{ g}\):\[n = \frac{0.300 \text{ g}}{12.01 \text{ g/mol}} = 0.02498 \text{ mol}.\]Next, calculate the heat evolved per mole of carbon:\[q_{\text{per mole}} = \frac{q_{\text{total}}}{n} = \frac{9847.19 \text{ J}}{0.02498 \text{ mol}} = 393,898.16 \text{ J/mol} \approx 393.9 \text{ kJ/mol}.\]

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Key Concepts

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

Heat Capacity
Heat capacity is an essential concept in calorimetry. It tells us how much heat a substance can absorb or release, perturbed by a change in temperature. The heat capacity ( C ) of an object is the amount of heat required to raise its temperature by one degree Kelvin or Celsius. In the case of a calorimeter, which is a device used to measure the amount of heat involved in a chemical reaction, understanding its heat capacity is crucial to determining accurate measurements.
The given problem involves a bomb calorimeter with a known heat capacity of 893 J/K. By knowing this value, along with the observed temperature change ( ΔT = 2.38^ ∘ C ), we can calculate how much heat the calorimeter itself absorbs from the reaction. This is done using the equation:
  • The formula: q calorimeter = Cbomb × ΔT
  • Substitute: 893 J/K × 2.38 K = 2124.34 J.
This result tells us precisely how much energy the calorimeter absorbed, helping us understand the full scope of the thermal energy exchange during the reaction.
Thermochemistry
Thermochemistry explores the heat energy involved in chemical reactions and physical transformations. It's an integral part of understanding calorimetry because it directly deals with how heat is exchanged in reactions like burning graphite to form CO 2 (g) . The core principle is based on the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred.
In the provided exercise, thermochemistry involves calculating how much heat energy is evolved during the combustion of carbon. By adding the heat absorbed by the water and the calorimeter, we determine the total heat evolved by the reaction ( q total = 9847.19 J). This is an example of an exothermic reaction, as it releases heat to its surroundings. This differentiation between heat evolved and absorbed clarifies how energy conservation plays out in practice. Calculating for the entire system, not merely individual parts, provides deeper insight into the net energy change.
Chemical Reactions
Chemical reactions, whether exothermic or endothermic, serve as fundamental catalysts for heat exchange in any calorimetry setup. They involve the breaking and forming of chemical bonds, translating chemical energy into heat energy, or vice versa.
In this exercise, we focus on the combustion of graphite (C graphite + O 2 (g) → CO 2 (g) ), an exothermic process. This reaction releases energy, which is then captured by the calorimeter and water, increasing their temperature. By understanding the stoichiometry of the reaction, the amount of heat released upon reacting a precise mass of carbon (0.300 g) can be calculated.
  • The number of moles of carbon reacting provides a scale for the reaction: 0.300 g of C corresponds to 0.02498 mol.
  • With this, the quantity of heat evolved per mole of carbon can be determined ( 393.9 kJ/mol).
Thus, this exercise not only elaborates on how chemical reactions produce heat but also showcases how calorimetry quantifies this energy release on a per mole basis, crucial for evaluating reaction efficiency and energy needs.

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Most popular questions from this chapter

A solar panel is pictured in the photo. When light shines on the panel, a small electric motor propels the car. What types of energy are involved in this setup?

A piece of lead with a mass of \(27.3 \mathrm{g}\) was heated to \(98.90^{\circ} \mathrm{C}\) and then dropped into \(15.0 \mathrm{g}\) of water at \(22.50^{\circ} \mathrm{C} .\) The final temperature was \(26.32^{\circ} \mathrm{C} .\) Calculate the specific heat capacity of lead from these data.

For each of the following, tell whether the process is exothermic or endothermic. (No calculations are required.) (a) \(\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) (b) \(2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(\mathrm{H}_{2} \mathrm{O}\left(\ell, 25^{\circ} \mathrm{C}\right) \longrightarrow \mathrm{H}_{2} \mathrm{O}\left(\ell, 15^{\circ} \mathrm{C}\right)\) (d) \(\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

The flashlight in the photo does not use batteries. Instead you move a lever, which turns a geared mechanism and results finally in light from the bulb. What type of energy is used to move the lever? What type or types of energy are produced?

Enthalpy changes for the following reactions can be determined experimentally: $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) \quad \Delta H^{\circ}=-91.8 \mathrm{kJ}$$ $$\begin{array}{r}4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\\\\Delta H^{\circ}=-906.2 \mathrm{kJ}\end{array}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad \Delta H^{\circ}=-241.8 \mathrm{kJ}$$ Use these values to determine the enthalpy change for the formation of \(\mathrm{NO}(\mathrm{g})\) from the elements (an enthalpy change that cannot be measured directly because the reaction is reactant-favored). $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g}) \quad \Delta H^{\circ}=?$$

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