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Chapter 5: Problem 57

A 2.200 -g sample of quinone \(\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}\right)\) is burned in a bomb calorimeter whose total heat capacity is \(7.854 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter increases from \(23.44^{\circ} \mathrm{C}\) to \(30.57^{\circ} \mathrm{C}\). What is the heat of combustion per gram of quinone? Per mole of quinone?

Short Answer

Expert verified
The heat of combustion per gram of quinone is \( 25.46 \frac{kJ}{g} \) and per mole of quinone is \( 2751 \, \frac{kJ}{mol} \).

Step by step solution

01

Calculate the heat released by the combustion of quinone

To do this, we will use the formula: \( q = C_{cal} \times \Delta T \) where q is the heat released, \( C_{cal} \) is the heat capacity of the calorimeter, and \( \Delta T \) is the change in temperature. Given, \( C_{cal} = 7.854 \frac{kJ}{^\circ C} \), the initial temperature \( T_i = 23.44 ^\circ C \) and the final temperature \( T_f = 30.57 ^\circ C \). First, let's find the change in temperature: \( \Delta T = T_f - T_i \)
02

Calculate the change in temperature

\( \Delta T = 30.57 ^\circ C - 23.44 ^\circ C = 7.13 ^\circ C \) Now that we have the change in temperature, let's calculate the heat released:
03

Calculate the heat released (q)

\(q = C_{cal} \times \Delta T = 7.854 \frac{kJ}{^\circ C} \times 7.13 ^\circ C \) \(q = 56.01 \, kJ\)
04

Calculate the heat of combustion per gram of quinone

Given that we have 2.200 g sample of quinone, we can find the heat of combustion per gram by dividing the heat released by the mass: \( q_{per\,gram} = \frac{q}{mass} = \frac{56.01 \, kJ}{2.200 \, g} \) \( q_{per\,gram} = 25.46 \frac{kJ}{g} \)
05

Calculate the molar mass of quinone

Quinone has the chemical formula \( \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2} \). We can calculate its molar mass by adding up the molar mass of each element: Molar mass of \(C = 12.01 \, g/mol\), of \(H = 1.01 \, g/mol\) and of \(O = 16.00 \, g/mol\). \( Molar\,mass\,of\,quinone = 6 \times 12.01 + 4 \times 1.01 + 2 \times 16.00 \) \( Molar\,mass\,of\,quinone = 108.10 \, g/mol \)
06

Calculate the heat of combustion per mole of quinone

Now that we have the molar mass of quinone, we can calculate the heat of combustion per mole: \( q_{per\,mole} = q_{per\,gram} \times Molar\,mass\,of\,quinone = 25.46 \frac{kJ}{g} \times 108.10 \, g/mol \) \( q_{per\,mole} = 2751 \, \frac{kJ}{mol} \) Therefore, the heat of combustion per gram of quinone is \( 25.46 \frac{kJ}{g} \) and per mole of quinone is \( 2751 \, \frac{kJ}{mol} \).

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