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

2. 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) 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

3. 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 kJ + 9 kJ = 569 kJ

4. 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

6. 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

7. 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!.

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

9. 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.

10. 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) H2O(g) ΔH = kJ
This is an exothermic reaction as ΔH is negative.
The reverse reaction would be endothermic.

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

12. 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.

13. 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.

14. Combustion calorimeter

15. 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.

16. 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.

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

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

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

20. 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) O2(g)  2CO2(g) H2O(l) H = kJ
C2H6(g) + 7/2O2(g)  2CO2(g) H2O(l) H = kJ
H2(g) /2O2(g)  H2O(l) H = -286 kJ
Consult your neighbor if necessary.

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

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

24. Standard Enthalpies of formation:

25. Thermodynamic Quantities of Selected Substances @ 298.15 K