Enthalpy and Calorimetry

Chapter 5 part 2

Enthalpy• H is heat under

constant pressure or

• H=qP

• H=E+PV• And therefore• ΔH= ΔE+P ΔV• ΔH=Hfinal-Hinitial

Example:

• When 1 mole of methane is burned at a constant pressure. 890kJ of energy is released as heat. Calculate the ΔH for a process in which a 5.8g sample of methane is burned at constant pressure.

Answer

• ΔH= -890kJ/mol • CH4 is methane, MM = 16 g/mol• 5.8g/16g mol-1 = 0.36 mol

• 0.36 mol CH4 x -890kJ/mol CH4 = -320kJ

Calorimetry

• The science of measuring heat.

• The device used to measure heat changes associated with chemical reaction is a calorimeter

Heat Capacity• Heat capacity of an

object is the heat absorbed by the change in temperature or

• C=(heat absorbed)/(ΔT)

• Specific heat capacity is the heat capacity per gram of a substance or

• Csp=C/gram• Molar heat capacity is

the heat capacity per mole of a substance or

• Cmol=C/mole

Calorimeters

• Constant pressure calorimeter

• Calculations (per gram)

• Csp* mass*ΔT= ΔHRemember:

• the specific heat and the mass are of the same object.

• The measurement is of the surrounding water, not the actual system so the sign is reversed.

Example:• When 1.00 L of 1.00M Ba(NO3)2 solution at

25.0 °C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25 °C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1 °C . Assuming that the calorimeter absorbs only a negligible amount of heat and the specific heat of the solution is 4.18J/°Cg and the density of the solution is 1.00g/mL, calculate the enthalpy change per mole of BaSO4 formed.

• Ba2+ (aq) + SO4 2- (aq)→BaSO4 (s)

• Mass of solution =2 liters = 2000 grams

• Csp= 4.18J/°Cg • ΔT=28.1-25.0 =3.1 °C• qsurroundings = 2.6 x104J

• qsystem= -2.6 x104J

Bomb Calorimeter

• This is also known as a constant volume calorimeter.

• Since it is under constant volume, but there is a change in temperature, the pressure is not constant.

Bomb Calorimeter

• Note: Since there is no change in volume, then no work is done.

• The constant volume calorimeter measures the system directly.

• The specific heat is of the calorimeter itself.

Example• In comparing potential fuels a bomb

calorimeter with a specific heat of 11.3kJ/ °C was employed. When a 1.50 g sample of methane was burned with an excess of oxygen in the calorimeter, the temperature increased by 7.3 °C . When a 1.15 g sample of hydrogen gas was burned with an excess of oxygen, the temperature increase was 14.3 °C. Calculate the energy of combustion per gram for these two fuels.

• Methane: released energy from 1.5 gram.

• =(11.3 kJ/g °C)(7.3 °C)

• =83kJ• Per gram• 83kj/1.5g=55kJ/g

• Hydrogen: released energy from 1.15 g.

• =(11.3 kJ/g °C)(14.3 °C)

• =162 kJ• Per gram• 162kJ/1.15 g= 141

kJ/g

Hess’s Law• Since enthalpy is a

state function, then• In going from a

particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.

Hess’s Law