Ch.6 - Thermochemistry WorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: A 0.3423 g sample of pentane, C5H12, was burned in a bomb calorimeter. The temperature of the calorimeter and the 1.000 kg of water contained therein rose from 20.22ºC to 22.82ºC. The heat capacity of

Solution: A 0.3423 g sample of pentane, C5H12, was burned in a bomb calorimeter. The temperature of the calorimeter and the 1.000 kg of water contained therein rose from 20.22ºC to 22.82ºC. The heat capacity of

Problem

A 0.3423 g sample of pentane, C5H12, was burned in a bomb calorimeter. The temperature of the calorimeter and the 1.000 kg of water contained therein rose from 20.22ºC to 22.82ºC. The heat capacity of the calorimeter is 2.21 kJ/ºC. The heat capacity of water = 4.184 J/gºC. How much heat was given off during combustion of the sample of pentane?

1) 8.8 kJ

2) -8.8 kJ

3) 16.6 J

4) 16.6 kJ

5) 3.1415 kJ

Solution

We’re being asked to calculate the heat of the reaction (qrxn) in a bomb calorimeter. This is given by the equation:



where qcalorimeter = heat absorbed by the calorimeter and qsolution = heat absorbed by the water/solution in the calorimeter. Expanding this, we have:



where Ccal = heat capacity of the calorimeter, m = mass of water (in grams), c = specific heat capacity of water, and ΔT = change in temperature = final T – initial T.


We can’t use maas of pentane to determine the heat of the reaction needed since the heat capacity of pentane isn’t given. This is why we used the heat absorbed by the calorimeter and the water instead.


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