Transcript Water Phase Change Graph

Temperature (K)
How much heat energy is required (at constant pressure) to convert
50g of ice at 100K to liquid water at 315K given the following data:
Cwater = 4.184 J/gK, Cice = 2.1 J/gK, Hfusion = 6.01 kJ/mol.
315 K
273 K
100 K
Energy (J)
Heating ice up to melting point: q = cice  mice  Tice = (2.1 J/gK)(50g)(273K-100K) = 18165 J
Converting ice to liquid water: qp = n  Hfus,water = (50g)(1 mol/18g)(6.01 kJ/mol) = 16.69 kJ
Heating up liquid water: q = cwater  mwater  Twater = (4.184J/gK)(50g)(315K-273K) = 8786 J
Total Heat Required = q1 + q2 + q3 = 18.165 kJ + 16.69 kJ + 8.786 kJ = 43.64 kJ
A certain piece of metal with a mass of 140.0 g is heated to 135°
C. The metal is then placed into a coffee cup calorimeter
containing 100.0 g of water initially at 25 ° C. Determine the
specific heat of the metal if the calorimeter and metal reach
thermal equilibrium at 37° C.
- q metal = q water
- 140.0g ·c (37° C - 135 ° C = 100.0 g ·4.18 J/gC · (37 ° C - 25 ° C )
c= 5016 J
140 g · 98 ° C
= .366 J/gC
Enthalpy- heat evolved or absorbed by the reaction
Thermochemical equation – balanced chemical
equations that show the associated enthalpy change.
Heat of reaction- enthalpy change that accompanies a
reaction
Standard heat of formation- the change in enthalpy for
the reaction that forms 1 mol of the compound
(usually reported at 298 K)
Guidelines for Thermochemical Equations :
1. Enthalpy is an extensive property.
The magnitude of ΔH is directly
proportional to the amount of reactant
consumed by the process.
How much heat is released when 4.50 g of methane gas is
burned in a constant – pressure system according to the equation
below?
CH4 (g) + 2O2(g)  CO2 (g) + 2 H2O (l) ΔH= -890 kJ
Guidelines for Thermochemical Equations :
2. The enthalpy change for a reaction is equal in
magnitude but opposite in sign to ΔH for the reverse
reaction.
CH4 (g) + 2O2(g)  CO2 (g) + 2 H2O (l) ΔH= -890 kJ
CO2 (g) + 2 H2O (l)  CH4 (g) + 2O2(g) ΔH= 890 kJ
Guidelines for Thermochemical Equations :
3. The enthalpy change for a reaction depends on the
state of the reactants and products.
Hess’s Law
Hess's Law of Constant Heat
Summation states that regardless of
the multiple stages or steps of a
reaction, the total enthalpy change for
the reaction is the sum of all changes.
Germain Henri Hess
Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Reaction
C + 2H2  CH4
C + O2  CO2
H2 + ½ O2  H2O
CH4  C + 2H2
Ho
-74.80 kJ
-393.50 kJ
-285.83 kJ
+74.80 kJ
Step #1: CH4 must appear on the reactant side, so we reverse
reaction #1 and change the sign on H.
Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Reaction
C + 2H2  CH4
C + O2  CO2
H2 + ½ O2  H2O
CH4  C + 2H2
C + O2  CO2
Ho
-74.80 kJ
-393.50 kJ
-285.83 kJ
+74.80 kJ
-393.50 kJ
Step #2: Keep reaction #2 unchanged, because CO2 belongs on
the product side
Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Reaction
C + 2H2  CH4
C + O2  CO2
H2 + ½ O2  H2O
CH4  C + 2H2
C + O2  CO2
2H2 + O2  2 H2O
Step #3: Multiply reaction #2 by 2
Ho
-74.80 kJ
-393.50 kJ
-285.83 kJ
+74.80 kJ
-393.50 kJ
-571.66 kJ
Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Reaction
C + 2H2  CH4
C + O2  CO2
H2 + ½ O2  H2O
Ho
-74.80 kJ
-393.50 kJ
-285.83 kJ
CH4  C + 2H2
+74.80 kJ
C + O2  CO2
-393.50 kJ
2H2 + O2  2 H2O
-571.66 kJ
CH4 + 2O2  CO2 + 2H2O
-890.36 kJ
Step #4: Sum up reaction and H
Hess’s Law
“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.”
H reaction   np H of (products)   nr H of (reactants)
Calculation of Heat of Reaction
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Hrxn =  n Hf(products) -  n  Hf(reactants)
Substance
CH4
O2
CO2
H2O
Hf
-74.80
0
-393.50
-285.83
kJ
kJ
kJ
kJ
Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]
Hrxn = -890.36 kJ