Chapter 25 – Energy from Chemical Reactions

Thermochemical Equations Remember: ΔH = Hproducts – Hreactants. The heat of reaction, ΔH, is negative when there is an overall release of energy (exothermic) and positive when heat is absorbed (endothermic). For any reaction, ΔH is directly proportional to the amount of substance. If twice as much of the reactants is used then the ΔH would also be doubled.

Thermochemical Equations When working with thermochemical equations, you need to be aware that: The coefficients of the reactants indicate the amounts, in mole, of each substance that react to give the specified heat change. States of reactants and products must be specified, since energy changes occur when solids are converted to liquids or liquids to gases, and vice versa. If a reaction occurs in reverse, it has the same magnitude of ΔH but the opposite sign.

Calculations using Thermochemical Equations Calculate the heat energy released when 50.00mL of 0.200M sodium hydroxide reacts with dilute excess hydrochloric acid. H+(aq) + OH-(aq)  H2O(l); ΔH=57.2kJ mol-1 From the equation: 1 mol of NaOH releases 57.2kJ of energy. n(NaOH) = 0.200 x 0.0500 = 0.0100mol Let 0.0100mol of NaOH release x kJ By proportion: x = 0.0100 57.2 1 x = 0.0100 x 57.2 = 0.52kJ Let’s do Q 1, 2 & 3 on p.405

The Connection between Energy and Temperature Change 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.

Specific Heat Capacity Question Example Calculate the energy required to heat 120mL of water for a cup of coffee to boiling point if the initial water temperature is 20.0°C. Since the density of water is 1g mL, the mass of 120mL is 120g. Energy (J) = SHC x mass (g) x ΔT (°C) Energy required to raise temperature of 120g of water by 1 degree if the temperature rose by 80 degrees Energy (J) = 4.184 x 120 x 80 = 40160J = 40.2kJ Let’s do Q 4 c), d) & e) on p.409

Measuring the Heat Released during a Reaction Enthalpy changes are measured directly using an instrument call a calorimeter. The first picture on the next page 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.

Calorimeters

Calorimeters 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.

Heat of Combustion The heat of combustion of a substance is defined as the energy released when a specified amount of the substance burns completely in oxygen. Heats of combustion are measured using a calorimeter.

Calorimeters 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

Example A bomb calorimeter was calibrated by passing 1.5A through the electric heater fro 50.1s at a potential difference of 6.05V. The temperature of the water in the calorimeter rose by 0.387°C. ΔH for the equation: CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) was determined by burning 8.58x10-4mol of methane gas in the calorimeter. The temperature rose from 20.241°C to 20.891°C. Step 1: Determine the calibration factor of the calorimeter using E = VIt Then divide by ΔT Overall… CF = E (VIt) ΔT Step 2: Calculate the energy change during the reaction. Energy change in calorimeter = calibration factor x ΔT Step 3: Calculate ΔH for the equation. ΔH = Energy change / mol

Example A bomb calorimeter was calibrated by passing 1.5A through the electric heater fro 50.1s at a potential difference of 6.05V. The temperature of the water in the calorimeter rose by 0.387°C. ΔH for the equation: CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) was determined by burning 8.58x10-4mol of methane gas in the calorimeter. The temperature rose from 20.241°C to 20.891°C. Step 1: Determine the calibration factor of the calorimeter using E = VIt E= 6.05V x 1.50A x 50.1s = 454.7J Since this energy made the temperature rise by 0.387°C, the energy required to raise the temperature by 1°C = 454.7/0.387 = 1175J°C-1. Step 2: Calculate the energy change during the reaction. ΔT = 20.891 – 20.241 = 0.650°C Energy change in calorimeter = calibration factor x ΔT = 1175x0.650 = 763.8J Step 3: Calculate ΔH for the equation. ΔH = Energy change / mol = 763.8 / (8.58x10-4) = 890210J ΔH = -890kJmol-1 (-ve because it indicates that energy has been released, explaining the increase in temperature. Let’s do Q 5 & 6 p.409