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Suppose you burned \(0.300 \mathrm{g}\) of \(\mathrm{C}(\mathrm{s})\) in an excess of \(\mathrm{O}_{2}(\mathrm{g})\) in a constant volume calorimeter to give \(\mathrm{CO}_{2}(\mathrm{g})\) \(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})\) The temperature of the calorimeter, which contained 775 g of water, increased from \(25.00^{\circ} \mathrm{C}\) to \(27.38^{\circ} \mathrm{C}\) The heat capacity of the bomb is \(893 \mathrm{J} / \mathrm{K}\). Calculate \(\Delta U\) per mole of carbon.

Short Answer

Expert verified
\(\Delta U = 393.52 \, \text{kJ/mol of C}\).

Step by step solution

01

Calculate the Temperature Change

The temperature change (\(\Delta T\)) can be determined by subtracting the initial temperature from the final temperature: \(\Delta T = 27.38^{\circ}C - 25.00^{\circ}C = 2.38^{\circ}C\).
02

Calculate the Heat Absorbed by Water

The heat absorbed by the water (\(q_{\text{water}}\)) is calculated using the formula \(q = mc\Delta T\). The specific heat capacity of water \(c\) is \(4.18\, \text{J/g} \cdot \text{°C}\). Therefore, \(q_{\text{water}} = 775\, \text{g} \times 4.18\, \text{J/g} \cdot \text{°C} \times 2.38\, \text{°C} = 7705.5\, \text{J}\).
03

Calculate the Heat Absorbed by the Bomb Calorimeter

The heat absorbed by the bomb calorimeter (\(q_{\text{bomb}}\)) is calculated using the formula \(q = C\Delta T\), where \(C\) is the heat capacity of the bomb, \(893\, \text{J/K}\). Therefore, \(q_{\text{bomb}} = 893\, \text{J/K} \times 2.38\, \text{K} = 2125.34\, \text{J}\).
04

Calculate Total Heat Released

The total heat released (\(q_{\text{total}}\)) is the sum of the heat absorbed by the water and the bomb: \(q_{\text{total}} = q_{\text{water}} + q_{\text{bomb}} = 7705.5\, \text{J} + 2125.34\, \text{J} = 9830.84\, \text{J}\).
05

Calculate Moles of Carbon Burned

Using the molar mass of carbon (\(12.01\, \text{g/mol}\)), calculate the moles of carbon burned: \(moles = \frac{0.300\, \text{g}}{12.01\, \text{g/mol}} = 0.02498\, \text{mol}\).
06

Calculate \(\Delta U\) per Mole of Carbon

The change in internal energy (\(\Delta U\)) per mole of carbon is calculated by dividing the total heat by the moles of carbon: \(\Delta U = \frac{9830.84\, \text{J}}{0.02498\, \text{mol}} = 393,518.64\, \text{J/mol}\). Since \(1\, \text{kJ} = 1000\, \text{J}\), we convert \(\Delta U\) to kJ/mol: \(\Delta U = 393.52\, \text{kJ/mol}\).

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Key Concepts

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

Enthalpy Change
In the realm of thermodynamics, the enthalpy change is a vital concept, especially in chemical reactions and processes. Though it can sound a bit daunting, it's essentially the total energy change when a chemical reaction occurs under constant pressure. However, in bomb calorimetry, we deal with constant volume, not pressure; hence, we focus on the internal energy change instead.

Enthalpy change, denoted as \( \Delta H \), is directly related to heat changes in reactions because \( \Delta H = q_p \), where \( q_p \) is the heat exchanged at constant pressure. In simple terms:
  • A negative \( \Delta H \) indicates an exothermic reaction wherein the system releases heat.
  • A positive \( \Delta H \) denotes an endothermic reaction in which the system absorbs heat.
When the reaction is performed in a bomb calorimeter, the \( \Delta H \) can be approximated through calculated total heat transfer.
Heat Capacity
Heat capacity is a key factor in calculating the heat change in calorimetry experiments. Simply put, it describes how much heat an object can absorb for each degree of temperature change. This property can vary with objects dependent on whether they have a constant volume or pressure.

In our context, we dealt with the heat capacity of both water and the bomb calorimeter:
  • Specific heat capacity of water is a constant \( 4.18 \, \text{J/g} \cdot \text{°C} \), allowing us to calculate the heat absorbed by the water easily.
  • The bomb calorimeter’s heat capacity was given as \( 893 \, \text{J/K} \), which was used to find how much heat the bomb absorbed.
Understanding this concept allows you to compute the heat involved when substances undergo temperature changes during reactions.
Combustion Reaction
A combustion reaction is a high-energy process where a substance rapidly combines with oxygen, resulting in the release of energy as heat and light. In this exercise, carbon combusts in the presence of oxygen to form carbon dioxide, as shown by the equation:
\(\text{C(s)} + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) \)

Combustion reactions are always exothermic, meaning they release energy. This release of energy results in the rise of the temperature of the surrounding environment or system, such as the water in the calorimeter, evidencing the change in temperature pressures during the reaction process.

These reactions are fundamental in calorimetry experiments as they enable the study of energy changes associated with chemical transformations.
Internal Energy Change
Internal energy change (\( \Delta U \) ) is another foundational concept in thermodynamics. It refers to the total change in the internal energy of a system as it undergoes a reaction or process. In bomb calorimetry, \( \Delta U \) provides an accurate measure of energy changes because reactions occur at constant volume.

The formula used in bomb calorimetry for finding internal energy change is based on the heat released or absorbed during the reaction:
  • The total heat transferred is first calculated by summing the heat absorbed by both water and the bomb.
  • This total heat is then divided by the number of moles of substance combusted to yield the \( \Delta U \) per mole.
In this exercise, we calculated \( \Delta U \) to understand the energy dynamics of carbon combustion in a calorimeter setting.

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Most popular questions from this chapter

In the reaction of two moles of gaseous hydrogen and one mole of gaseous oxygen to form two moles of gaseous water vapor, two moles of products are formed from 3 moles of reactants. If this reaction is done at \(\left.1.0 \text { atm pressure (and at } 0^{\circ} \mathrm{C}\right),\) the volume is reduced by \(22.4 \mathrm{L}\) (a) In this reaction, how much work is done on the system \(\left(\mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{O}\right)\) by the surroundings? (b) The enthalpy change for this reaction is \(-483.6 \mathrm{kJ}\) Use this value, along with the answer to (a), to calculate \(\Delta_{r} U\), the change in internal energy in the system.

You determine that 187 J of energy as heat is required to raise the temperature of \(93.45 \mathrm{g}\) of silver from \(18.5^{\circ} \mathrm{C}\) to \(27.0^{\circ} \mathrm{C} .\) What is the specific heat capacity of silver?

Insoluble \(\mathrm{PbBr}_{2}(\mathrm{s})\) precipitates when solutions of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) and \(\mathrm{NaBr}(\mathrm{aq})\) are mixed. \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaBr}(\mathrm{aq}) \rightarrow \mathrm{PbBr}_{2}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})\) $$ \Delta_{\mathrm{r}} H^{\circ}=? $$ To measure the enthalpy change, \(200 .\) mL of \(0.75 \mathrm{M}\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) and \(200 . \mathrm{mL}\) of \(1.5 \mathrm{M} \mathrm{NaBr}(\mathrm{aq})\) are mixed in a coffee-cup calorimeter. The temperature of the mixture rises by \(2.44^{\circ} \mathrm{C} .\) Calculate the enthalpy change for the precipitation of \(\mathrm{PbBr}_{2}(\mathrm{s}),\) in \(\mathrm{k} \mathrm{J} / \mathrm{mol}\). (Assume the density of the solution is \(1.0 \mathrm{g} / \mathrm{mL},\) and its specific heat capacity is \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\).

An "ice calorimeter" can be used to determine the specific heat capacity of a metal. A piece of hot metal is dropped onto a weighed quantity of ice. The energy transferred from the metal to the ice can be determined from the amount of ice melted. Suppose you heated a 50.0 -g piece of silver to \(99.8^{\circ} \mathrm{C}\) and then dropped it onto ice. When the metal's temperature had dropped to \(0.0^{\circ} \mathrm{C},\) it is found that \(3.54 \mathrm{g}\) of ice had melted. What is the specific heat capacity of silver?

What determines the directionality of energy transfer as heat?

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