•The amount of energy required to raise the temperature of one gram of a substance by 1°C is called the specific heat capacity of that substance.
The higher the specific heat capacity, the more effectively a material will store heat energy.
•Heat change gained or lost by a substance during a chemical reaction can be calculated by:
E = m x c x ΔT
where E is Energy (J); m is Mass (g); c is specific heat capacity (Jg-1oC-1); T is temperature (oC), Δ means change in.
•Calculate the energy required to heat 120mL of water for a cup of coffee to boiling point if the initial water temperatuer is 20.0°C.
Since the density of water is 1g mL, the mass of 120mL is 120g.
E = m x c x ΔT
Energy required to raise temperature of 120g of water by 1 degree = SHC x mass
4.184 x 120 = 502.0J
Since temperature rises by 80 degrees, the total energy required = J x ΔT
= 502.0 x 80 = 40160J = 40.2kJ
Measuring heat released during a reaction
•Enthalpy changes are measured directly using an instrument call a calorimeter.
•The picture on the right shows the components of a bomb calorimeter used for reactions that involved gaseous reactants or products.
•The reaction vessel is designed to withstand high pressures created during the reaction.
•The second picture, is a calorimeter used for reactions in aqueous solutions.
•Both calorimeters are insulated to reduce loss or gain of energy to or from the outside environment.
•When a reaction takes place in a calorimeter, the heat change causes a rise or fall in the temperature of the contents of the calorimeter.
•Before the calorimeter can be of use, it must first be determined how much energy is required to change the temperature within the calorimeter by 1°C.
•This is known as the calibration factor of the calorimeter.
•The calorimeter is calibrated by using an electric heater to release a known quantity of thermal energy and measuring the resultant rise in temperature.
•The thermal energy released when an electric current passes through the heater can be calculated from the formula:
Energy = voltage (volts) x current(amps) x time (seconds) OR;
E = VIt
Alternatively Energy can be calculated using:
E = n xΔH for a known ΔH value
THE HEAT OF COMBUSTION
•The heat of combustion of a substance is defined as the energy released when a specified amount (eg. 1 mol, 1g, 1 L) of the substance burns completely in oxygen.
•Heats of combustion are measured using a calorimeter.
•ΔHc are given for substances in the data booklet (See below).
Diagram By Akshat Goel - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=18702902
Bomb calorimeter diagram By Lisdavid89 - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=22537546
The calibration factor
•The calibration factor is determined using:
cf = Energy added where cf = calibration factor
This can be used to find the energy of reaction:
Energy reaction = cf x ΔT reaction
FINDING ΔH FROM CALORIMETRY
ΔH can be determined using the following relationship:
ΔH = E reaction
Where n1= amount of reactant in the equation (mole ratio) and n2 = amount of mole used (given in the question).
Eg. CH4(g) + 2O2 (g) --> CO2 (g) + 2H2O (l)
Given that 0.1 mol of methane is burnt,
n1= 1 (mole ratio in equation), n2= 0.1 given in question.
•A bomb calorimeter was calibrated by passing 1.5A through the electric heater for 50.0s at a potential difference of 5.43V. The temperature of the water in the calorimeter rose by 0.412°C.
ΔH for the equation: CH4(g) + 2O2(g) à CO2(g) + 2H2O(l)
was determined by burning 9.21x10-4mol of methane gas in the calorimeter. The temperature rose from 20.40°C to 21.23°C.
Step 1: Determine the calibration factor of the calorimeter using E = VIt
E= 5.43V x 1.50A x 50.)s = 406.94J
Since this energy made the temperature rise by 0.387°C, the energy required to raise the temperature by 1°C = 406.94/0.412 = 987.71J°C-1.
Step 2: Calculate the energy change during the reaction.
ΔT = 21.23 – 20.40 = 0.83°C
Energy change in calorimeter = calibration factor x ΔT
= 987.71x0.83 = 819.8J
Step 3: Calculate ΔH for the equation.
Note- mole ratio of 1, n1 = 1
ΔH = Energy change / mol
= 819.8 / (9.21x10-4) = 890119J
ΔH = -890kJmol-1 (-ve because it indicates that energy has been released, explaining the increase in temperature.