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The following substances undergo complete combustion in a bomb calorimeter. The calorimeter assembly has a heat capacity of \(5.136 \mathrm{kJ} /^{\circ} \mathrm{C} .\) In each case, what is the final temperature if the initial water temperature is \(22.43^{\circ} \mathrm{C} ?\) \(\begin{array}{lllll}\text { (a) } 0.3268 & \text { g caffeine, } & \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{N}_{4} & \text { (heat of }\end{array}\) combustion \(=-1014.2 \mathrm{kcal} / \mathrm{mol} \text { caffeine })\) (b) \(1.35 \mathrm{mL}\) of methyl ethyl ketone, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}(1)\) \(d=0.805 \mathrm{g} / \mathrm{mL}\) (heat of combustion \(=-2444 \mathrm{kJ} / \mathrm{mol}\) methyl ethyl ketone).

Short Answer

Expert verified
The final temperature of the water after the combustion of caffeine is 22.747°C, and for methyl ethyl ketone it is 29.575°C.

Step by step solution

01

Calculate number of moles of the first substance

Calculate the number of moles (n₁) for caffeine as the molecular weight of caffeine is 194.19 g/mol.\n\n\[ n₁ = 0.3268/194.19 = 0.001682 mol \]
02

Calculate change in temperature for the first substance

Next, the change in temperature (∆t₁) due to combustion of caffeine can be calculated using the relation ∆t = - ( ∆Hc ⋅ n ) / C.\n\n\[ ∆t₁ = - ( -1014.2 × 0.001682 ) / 5.136 = 0.317°C \]
03

Final temperature for the first substance

Now we can find the final temperature of the water (T₁) for caffeine combustion, by adding the initial temperature to ∆t₁.\n\n\[ T₁ = 22.43 + 0.317 = 22.747°C \]
04

Calculate number of moles of the second substance

Now, calculate the number of moles (n₂) for methyl ethyl ketone (C₄H₈O). The molecular weight of C₄H₈O is 72.105 g/mol, and the given volume is 1.35 mL with a density of 0.805 g/mL, from which we get the mass.\n\n\[ m = 1.35 × 0.805 = 1.08675 g \]\n\n\[ n₂ = 1.08675/72.105 = 0.015064 mol \]
05

Calculate change in temperature for the second substance

Now, the change in temperature (∆t₂) due to combustion of methyl ethyl ketone can be calculated in a similar manner as in Step 2. However, we use 2444 kJ/mol instead of kcal/mol and also insert the heat of combustion with a negative value.\n\n\[ ∆t₂ = - ( -2444 × 0.015064 ) / 5.136 = 7.145°C \]
06

Final temperature for the second substance

Finally, the final temperature of the water upon combustion of methyl ethyl ketone (T₂) can be calculated by adding the initial temperature to ∆t₂.\n\n\[ T₂ = 22.43 + 7.145 = 29.575°C \]

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Key Concepts

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

Combustion Reactions
Combustion reactions are chemical processes where a substance combines with oxygen, releasing energy in the form of heat and sometimes light. We commonly refer to these as burning. These reactions are exothermic, meaning they produce heat as they proceed. Combustion is vital in energy production, such as in car engines, power plants, and even for heating our homes.
  • The reaction generally involves hydrocarbons or other organic compounds.
  • Oxygen from the air is the typical reactant that fuels the combustion process.
  • The products typically include carbon dioxide (CO₂) and water (H₂O).
In a bomb calorimeter, the combustion reaction is contained in a sealed vessel, allowing precise measurement of the heat produced. This setup provides an accurate assessment of how much energy a specific substance releases upon combustion. Understanding combustion reactions is crucial for calculating energy changes in chemical processes.
Heat Capacity
Heat capacity is an important concept in thermodynamics and refers to the amount of heat required to change the temperature of a system by one degree Celsius. This property tells us how well a system can hold and absorb heat. Specifically, in our exercise, we use the heat capacity of the calorimeter to determine how much the temperature changes during a reaction.
  • Heat capacity can be expressed in units of joules per degree Celsius (J/°C) or kilojoules per degree Celsius (kJ/°C).
  • A higher heat capacity means that more heat is required to change the temperature of the system.
  • In a bomb calorimeter, the heat capacity often includes the calorimeter's components and any water surrounding the reaction vessel.
Knowing the heat capacity allows us to use temperature change measurements to calculate the energy change for the reaction. This concept is essential in calorimetry experiments to yield accurate results.
Temperature Change
Temperature change is a key indicator of the energy exchanged in a chemical reaction. In the context of bomb calorimetry, the temperature change of the water in the calorimeter indicates how much energy is released during combustion.
  • The formula to calculate temperature change (\( \Delta t \)) is: \( \Delta t = \frac{- ( \Delta H_c \cdot n )}{C} \), where \( \Delta H_c \) is the heat of combustion, \( n \) is the number of moles, and \( C \) is the heat capacity of the calorimeter.
  • A positive \( \Delta t \) indicates that the temperature rises as the reaction proceeds, signifying an exothermic reaction.
  • By knowing the initial and final temperatures, one can deduce how much heat was released in physical terms during the reaction.
Temperature change allows chemists to quantify the energy produced by various substances in a controlled environment, thereby understanding the reaction's efficiency and energy output.
Caffeine Combustion
Caffeine combustion refers to the reaction where caffeine (\( \text{C}_8 \text{H}_{10} \text{O}_2 \text{N}_4 \)) is burned or oxidized in the presence of oxygen, typically done in a bomb calorimeter to measure the energy released. In our exercise, we learned how to calculate the temperature change resulting from caffeine combustion.
  • First, determine the number of moles of caffeine using its molecular weight, calculated as \( \frac{0.3268}{194.19} \approx 0.001682 \text{mol}. \)
  • The heat of combustion (\(-1014.2 \text{kcal/mol} \)) tells us how much energy is released per mole of substance.
  • By substituting values into the formula for \( \Delta t \), we find the temperature change due to caffeine combustion is 0.317°C, signifying a rise in temperature.
Understanding caffeine combustion is important for estimating how substances like caffeine release energy under controlled conditions. This has implications for both food chemistry and energetics.
Methyl Ethyl Ketone Combustion
Methyl ethyl ketone (MEK) is a flammable liquid often used as a solvent. Its combustion in a bomb calorimeter provides insight into its energetic properties. In our exercise, we calculated the temperature change resulting from the combustion of MEK.
  • First, the mass of MEK is determined by multiplying its volume (1.35 mL) by its density (0.805 g/mL), resulting in 1.08675 g.
  • The number of moles is then calculated using its molecular weight (72.105 g/mol), yielding \( \frac{1.08675}{72.105} \approx 0.015064 \text{mol}. \)
  • Using the heat of combustion (\(-2444 \text{kJ/mol} \)), we find a temperature rise of 7.145°C, indicating significant energy release.
Studying MEK combustions helps understand solvent energetics, and it has practical applications in industries involving chemical manufacturing and processing.

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Most popular questions from this chapter

The combustion of methane gas, the principal constituent of natural gas, is represented by the equation $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ} &=-890.3 \mathrm{kJ} \end{aligned}$$ (a) What mass of methane, in kilograms, must be burned to liberate \(2.80 \times 10^{7} \mathrm{kJ}\) of heat? (b) What quantity of heat, in kilojoules, is liberated in the complete combustion of \(1.65 \times 10^{4} \mathrm{L}\) of \(\mathrm{CH}_{4}(\mathrm{g})\) measured at \(18.6^{\circ} \mathrm{C}\) and \(768 \mathrm{mmHg} ?\) (c) If the quantity of heat calculated in part (b) could be transferred with \(100 \%\) efficiency to water, what volume of water, in liters, could be heated from 8.8 to \(60.0^{\circ} \mathrm{C}\) as a result?

A 465 g chunk of iron is removed from an oven and plunged into \(375 \mathrm{g}\) water in an insulated container. The temperature of the water increases from 26 to \(87^{\circ} \mathrm{C}\). If the specific heat of iron is \(0.449 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1},\) what must have been the original temperature of the iron?

Thermite mixtures are used for certain types of welding, and the thermite reaction is highly exothermic. $$\begin{array}{r} \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Fe}(\mathrm{s}) \\ \Delta H^{\circ}=-852 \mathrm{kJ} \end{array}$$ \(1.00 \mathrm{mol}\) of granular \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(2.00 \mathrm{mol}\) of granular Al are mixed at room temperature \(\left(25^{\circ} \mathrm{C}\right),\) and a reaction is initiated. The liberated heat is retained within the products, whose combined specific heat over a broad temperature range is about \(0.8 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} .\) (The melting point of iron is \(1530^{\circ} \mathrm{C} .\) ) Show that the quantity of heat liberated is more than sufficient to raise the temperature of the products to the melting point of iron.

The metabolism of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) yields \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) as products. Heat released in the process is converted to useful work with about \(70 \%\) efficiency. Calculate the mass of glucose metabolized by a \(58.0 \mathrm{kg}\) person in climbing a mountain with an elevation gain of \(1450 \mathrm{m}\). Assume that the work performed in the climb is about four times that required to simply lift \(58.0 \mathrm{kg}\) by \(1450 \mathrm{m} \cdot\left(\Delta H_{\mathrm{f}}^{2} \text { of } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}) \text { is }-1273.3 \mathrm{kJ} / \mathrm{mol} .\right)\)

A coffee-cup calorimeter contains \(100.0 \mathrm{mL}\) of \(0.300 \mathrm{M}\) HCl at \(20.3^{\circ} \mathrm{C}\). When \(1.82 \mathrm{g} \mathrm{Zn}(\mathrm{s})\) is added, the temperature rises to \(30.5^{\circ} \mathrm{C}\). What is the heat of reaction per mol Zn? Make the same assumptions as in Example \(7-4,\) and also assume that there is no heat lost with the \(\mathrm{H}_{2}(\mathrm{g})\) that escapes. $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$$

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