Ch. 16: Energy and Chemical Change Sec. 16.3: Thermochemical Equations

Objectives Write thermochemical equation for chemical reactions and other processes. Describe how energy is lost or gained during changes of state. Calculate the heat absorbed or released in a chemical reaction.

Review The change in energy is an important part of chemical reactions so chemists include ΔH as part of the chemical equation.

Thermochemical Equations A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products and the energy change. The energy change is usually expressed as the change in enthalpy, ΔH.

Thermochemical Equations A subscript of ΔH will often give you information about the type of reaction or process taking place. For example, ΔHcomb is the change in enthalpy for a combustion reaction or, simply, the enthalpy (heat) of combustion.

Thermochemical Equations C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔHcomb = -2808 kJ/mol This is the thermochemical equation for the combustion of glucose. The enthalpy of combustion is -2808 kJ/mol.

Thermochemical Equations The enthalpy of combustion (ΔHcomb) of a substance is defined as the enthalpy change for the complete burning of one mole of the substance. That means, 2808 kJ of heat are released for every mole of glucose that is oxidized (or combusts). Two moles would release 5616 kJ: 2 mol glucose x -2808 kJ = -5616 kJ 1 mol The zero superscript tells you that the reactions were carried out under standard conditions. Standard conditions are one atmosphere of pressure and 298 K (25 degrees C).

Standard Enthalpies Standard enthalpy changes have the symbol ΔHo. Standard conditions in thermochemistry are 1 atm pressure and 25 0C. **DO NOT CONFUSE THESE WITH THE STP CONDITIONS OF THE GAS LAWS.

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) Calculations How much heat is evolved when 54.0 g of glucose (C6H12O6) is burned according to this equation? C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔHcomb = -2808 kJ/mol Since the ΔHcomb given is for 1 mole of glucose, you must first determine the # of moles you have. 54.0 g x 1 mole = 0.300 moles glucose 180 g

Calculations (cont.) You can now use the ΔHcomb value as a conversion factor: 0.300 mol glucose x -2808 kJ = -842 kJ 1 mole

Practice Problems How much heat will be released when 6.44 g of sulfur reacts with O2 according to this equation: 2S + 3O2  2SO3 ΔH0 = -395.7 kJ/mol How much heat will be released when 11.8 g of iron react with O2 according to the equation: 3Fe + 2O2  Fe3O4 ΔH0 = -373.49 kJ/mol

Changes of State The heat required to vaporize one mole of a liquid is called its molar enthalpy (heat) of vaporization (ΔHvap). The heat required to melt one mole of a solid substance is called its molar enthalpy (heat) of fusion (ΔHfus). Both phase changes are endothermic & ΔH has a positive value. Because vaporizing a liquid and melting a solid are endothermic processes, their ΔH values are positive. These are standard values that can be looked up.

Changes of State The vaporization of water and the melting of ice can be described by the following equations: H2O(l) → H2O(g) ΔHvap = 40.7 kJ/mol One mole of water requires 40.7 kJ to vaporize. H2O(s) → H2O(l) ΔHfus = 6.01 kJ/mol One mole of ice requires 6.01 kJ to melt. The 1st equation indicates that 40.7 kJ of energy is absorbed when one mole of water is converted to one mole of water vapor. The 2nd equation shows that when one mole of ice melts to form one mole of liquid water, 6.01 kJ of energy is absorbed. What happens in the reverse process?

ΔHcond = - ΔHvap = -40.7 kJ/mol ΔHsolid = - ΔHfus = -6.01 kJ/mol Changes of State The same amounts of energy are released in the reverse processes (condensation and solidification (freezing)) as are absorbed in the processes of vaporization and melting. Therefore, they have the same numerical values but are opposite in sign. ΔHcond = - ΔHvap = -40.7 kJ/mol ΔHsolid = - ΔHfus = -6.01 kJ/mol

Practice Problems How much heat is released when 275 g of ammonia gas condenses at its boiling point? (ΔHvap = 23.3 kJ/mol) If water at 00 C releases 52.9 J as it freezes, what is the mass of the water? (ΔHfus= 6.01 kJ/mol) How much heat is required to melt 25 g of ice at its MP? (ΔHfus= 6.01 kJ/mol)

Combination Problems At times, you will need to calculate the amount of heat that is absorbed or released when a temperature change AND a phase change occur in sequence. Recall, the heat involved in a temperature change is calculated by using: ΔH = mCΔT. NOTE: In this expression, ΔH is found in joules or cal. Heat involved in a phase change is calculated using dimensional analysis. NOTE: ΔH will be in kJ.

Example How much heat is released when 37.5 g of water that is at 20.0 0C freezes? (ΔHfus= 6.01 kJ/mol) Water cannot freeze at 20.0 0C. It must be at 0 0C. Therefore, we must first determine how much energy is released when it is cooled from 20.0 0C to 0 0C. ΔH = mCΔT = (37.5 g)(4.184 J/g 0C)(- 20.0 0C) = - 3138 J = -3.14 kJ

Example How much heat is released when 37.5 g of water that is at 20.0 0C freezes? (ΔHfus= 6.01 kJ/mol) -3.14 kJ of heat are released when the sample is cooled to 0 0C. Now we must calculate the energy released when the entire sample freezes. Recall that ΔHsolid= -ΔHfus= -6.01kJ/mol 37.5 g x 1 mole x -6.01 kJ = -12.5 kJ 18 g 1 mole

Example How much heat is released when 37.5 g of water that is at 20.0 0C freezes? (ΔHfus= 6.01 kJ/mol) -3.14 kJ of heat are released when the sample is cooled to 0 0C. -12.5 kJ of heat are released when the sample freezes. The total heat released is -3.14 + -12.5 or -15.6 kJ.

Use Cw = 4.184 J/g0C; ΔHfus = 6.01 kJ/mol; ΔHvap = 40.7 kJ/mol. Practice Problems Use Cw = 4.184 J/g0C; ΔHfus = 6.01 kJ/mol; ΔHvap = 40.7 kJ/mol. How much heat is needed to melt 8 g of ice at 0 0C to water at 15 0C? How much heat is needed to change 28.0 g of water at 60.0 0C to steam?