|---> ∆Hvap ---> ---> ---> -->| How many joules are required to heat 200 grams of water from 25 0C to 125 0C? |---> ∆Hvap ---> ---> ---> -->| |<--- <--- <--- <--- ∆Hcond <---| ∆Hfus ∆Hsolid

How many joules are required to heat 200 grams of water from 25 0C to 125 0C? The heat capacity of steam is 1.7 J / g . 0C Temp change state of water formula Heat change (KJ) 1 KJ=1000J 25 0C to 100 0C Liquid q1 = m x C x ∆T   100 0C LiquidGas Q2=m ∆Hvap 100 0C to 125 0C Gas q3 = m x C x ∆T Total 523 kJ

q1 = m x C x ∆T = (200 g) x (4.184 J/(g °C)) x (100 °C – 25 °C) = (200 g) x (4.184 J/(g °C)) x ( 75 °C) q1 = 6 x 105 J or 60 kJ   q2 = ∆Hvap x moles q = (40.7 kJ/mole)((200 g H2O)(1mole/18.02 g)) q2 = 500 kJ q3 = m x C x ∆T = (200 g) x (1.7 J/(g °C)) x (125 °C – 100 °C) = (200 g) x (1.7 J/(g °C)) x ( 25 °C) q3 = 9 x 103 J or 9 kJ q = q1+q2+q3 q = 60 kJ + 500 kJ + 9 kJ = 569 kJ

How many joules are given off when 120 grams of water are cooled from 25 0C to -250C? The heat capacity of ice is 2.1 J / g . 0C. Temp change state of water formula Heat change (KJ) 25 0C to 00C Liquid q1 = m x C x ∆T    0 C   LS Q=m ∆H  0 to -25 Solid   q=m x c x ∆T Total -59 kJ

q1 = m x C x ∆T = (120 g) x (4.184 J/(g °C)) x (0 °C – 25 °C) = (120 g) x (4.184 J/(g °C)) x ( -25 °C) q1 = -1.3 x 104 J or -13 kJ   q2 = ∆Hsol x moles q = (-6.01 kJ/mole)((120 g H2O)(1mole/18.02 g)) q2 = -40 kJ q3 = m x C x ∆T = (120 g) x (2.1 J/(g °C)) x (-25 °C – 0 °C) = (120 g) x (2.1 J/(g °C)) x ( -25 °C) q3 = -6.3 x 103 J or -6.3 kJ q = q1+q2+q3 q = -13 kJ -40 kJ -6.3 kJ = -59 kJ

Steps for solving: a) Draw a phase change diagram and indicate which areas are involved b) Write all the appropriate equations c) Solve the problem and add up energy – make sure the units are all the same!.

Enthalpy of Reaction The quantity of energy transferred as heat during a chemical reaction.

In an exothermic reaction the enthalpy change is negative – meaning energy is released from the system as heat. In an endothermic reaction the enthalpy change is positive – meaning energy is absorbed into the system as heat.

Thermochemical equations A thermochemical equation is an equation that includes the quantity of energy released or absorbed as heat during the reaction. Example 2H2(g) + O2(g) ------ 2H2O(g) ΔH = - 483.6kJ This is an exothermic reaction as ΔH is negative. The reverse reaction would be endothermic.

The reverse reaction would be endothermic. 2H2O(g) ---- 2H2(g) + O2(g) ΔH = +483.6kJ

Enthalpy of Formation The enthalpy change that occurs when one mole of a compound is formed from its elements in their standard state at 25 degrees and 1 atm. These standard enthalpies are given in the Appendix Table A-14. The majority of enthalpies of formation are negative as the products formed are more stable and of lower energy.

Enthalpy of Combustion Combustion reactions form a large amount of energy in the form of heat and light when a substance is combined with oxygen. The enthalpy of combustion is the amount of heat energy produced when one mole of a substance burns in oxygen.

Combustion calorimeter

Hess’s Law Start Finish A State Function: Path independent. Both lines accomplished the same result, they went from start to finish. Net result = same.

Hess’s Law states that the energy change in an overall chemical reaction is equal to the sum of the energy changes in the individual reactions comprising it.

and.. the H values must be treated accordingly. Determine the heat of reaction for the reaction: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ Hint: The three reactions must be algebraically manipulated to sum up to the desired reaction. and.. the H values must be treated accordingly.

Found in more than one place, SKIP IT (its hard). Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ 4NH3  2N2 + 6H2 H = +183.6 kJ NH3: Reverse and x 2 Found in more than one place, SKIP IT (its hard). O2 : NO: x2 2N2 + 2O2  4NO H = 361.2 kJ H2O: x3 6H2 + 3O2  6H2O H = -1451.1 kJ

Found in more than one place, SKIP IT. Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) 4NH3  2N2 + 6H2 H = +183.6 kJ NH3: Reverse and x2 Found in more than one place, SKIP IT. O2 : NO: x2 2N2 + 2O2  4NO H = 361.2 kJ H2O: x3 6H2 + 3O2  6H2O H = -1451.1 kJ Cancel terms and take sum. + 5O2 + 6H2O H = -906.3 kJ 4NH3  4NO Is the reaction endothermic or exothermic?

Consult your neighbor if necessary. Determine the heat of reaction for the reaction: C2H4(g) + H2(g)  C2H6(g) Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ Consult your neighbor if necessary.

Determine the heat of reaction for the reaction: Goal: C2H4(g) + H2(g)  C2H6(g) H = ? Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H4(g) :use 1 as is C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ H2(g) :# 3 as is H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H6(g) : rev #2 2CO2(g) + 3H2O(l)  C2H6(g) + 7/2O2(g) H = +1550 kJ C2H4(g) + H2(g)  C2H6(g) H = -137 kJ

enthalpy is a state function and is path independent. Summary: enthalpy is a state function and is path independent.

Standard Enthalpies of formation:

Thermodynamic Quantities of Selected Substances @ 298.15 K