Ch 15 Energy and Chemical Change Sec 15

Ch. 15: Energy and Chemical Change Sec. 15. 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 enthalpy change for the complete burning of one mole of a substance or, simply, the enthalpy (heat) of combustion.

Thermochemical Equations C 6 H 12 O 6(s) + 6 O 2(g) → 6 CO 2(g) + 6 H 2 O(l) ΔHcomb = -2808 k. J/mol This is thermochemical equation for the combustion of glucose. The enthalpy of combustion is -2808 k. J/mol.

Thermochemical Equations Since the enthalpy of combustion (ΔHcomb) of a substance is defined as the enthalpy change for the complete burning of one mole of the substance, this means that 2808 k. J of heat are released for every mole of glucose that is oxidized (or combusts). Two moles, therefore, would release 5616 k. J: 2 mol glucose x -2808 k. J = -5616 k. J 1 mol

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

Calculations How much heat is evolved when 54. 0 g of glucose (C 6 H 12 O 6) is burned according to this equation? C 6 H 12 O 6(s) + 6 O 2(g) → 6 CO 2(g) + 6 H 2 O(l) ΔHcomb = -2808 k. J/mol n 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. ) n You can now use the ΔHcomb value as a conversion factor: 0. 300 mol glucose x -2808 k. J = -842 k. J 1 mole

Practice Problems How much heat will be released when 6. 44 g of sulfur reacts with O 2 according to this equation: 2 S + 3 O 2 2 SO 3 ΔH 0 = -395. 7 k. J/mol How much heat will be released when 11. 8 g of iron react with O 2 according to the equation: 3 Fe + 2 O 2 Fe 3 O 4 ΔH 0 = -373. 49 k. J/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.

Changes of State The vaporization of water and the melting of ice can be described by the following equations: H 2 O(l) → H 2 O(g) ΔHvap = 40. 7 k. J/mol n One mole of water requires 40. 7 k. J to vaporize. H 2 O(s) → H 2 O(l) ΔHfus = 6. 01 k. J/mol One mole of ice requires 6. 01 k. J to melt.

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 k. J/mol ΔHsolid = - ΔHfus = -6. 01 k. J/mol

Practice Problems How much heat is released when 275 g of ammonia gas condenses at its boiling point? (ΔHvap = 23. 3 k. J/mol) If water at 00 C releases 52. 9 J as it freezes, what is the mass of the water? (ΔHfus= 6. 01 k. J/mol) How much heat is required to melt 25 g of ice at its MP? (ΔHfus= 6. 01 k. J/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. n n Recall, the heat involved in a temperature change is calculated by using: ΔH = m. CΔT. 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 k. J.

Example How much heat is released when 37. 5 g of water that is at 20. 0 0 C freezes? (ΔHfus= 6. 01 k. J/mol) Water cannot freeze at 20. 0 0 C. It must be at 0 0 C. Therefore, we must first determine how much energy is released when it is cooled from 20. 0 0 C to 0 0 C. ΔH = m. CΔT = (37. 5 g)(4. 184 J/g 0 C)(- 20. 0 0 C) = - 3138 J = -3. 14 k. J n

Example How much heat is released when 37. 5 g of water that is at 20. 0 0 C freezes? (ΔHfus= 6. 01 k. J/mol) -3. 14 k. J of heat are released when the sample is cooled to 0 0 C. n Now we must calculate the energy released when the entire sample freezes. Recall that ΔHsolid= -ΔHfus= -6. 01 k. J/mol 37. 5 g x 1 mole x -6. 01 k. J = -12. 5 k. J 18 g 1 mole n

Example How much heat is released when 37. 5 g of water that is at 20. 0 0 C freezes? (ΔHfus= 6. 01 k. J/mol) n n n -3. 14 k. J of heat are released when the sample is cooled to 0 0 C. -12. 5 k. J of heat are released when the sample freezes. The total heat released is -3. 14 + -12. 5 or 15. 6 k. J.

Practice Problems Use Cw = 4. 184 J/g 0 C; ΔHfus = 6. 01 k. J/mol; ΔHvap = 40. 7 k. J/mol. How much heat is needed to melt 8 g of ice at 0 0 C to water at 15 0 C? How much heat is needed to change 28. 0 g of water at 60. 0 0 C to steam?