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Chapter 6: THERMOCHEMISTRY
Thermochemistry- Study of the relationship between chemical
reactions and energy changes involving heat.
Kinetic Energy- Energy an object has because of its motion.
K.E. = (1/2)mv2
v = velocity(m/s)
m = mass of moving object in kilograms
Potential Energy – Energy an object has due to its position OR
energy that is stored for example in a chemical bond. Chemical
reaction is an example of potential energy.
P.E. = mgh
h = height from ground
g = gravitational acceleration (9.8 m/s2)
m = mass of object in kilograms
Electrostatic Energy – Form of potential energy that is derived
from the interaction of charged particles or objects
E = (KQ1Q2)/d ; Q is charge, d is distance between charges, K is
proportionality constant.
Units of Energy
SI unit of energy is the joule.
The metric units of the joule is kg (m2/s2)
Ex. 2kg object moving at 1m/s has (1/2)(2kg)(1m/s)2 =
1kg(m2/s2) = 1joule
Another unit of energy is calorie; 1 calorie = 4.184 joules, the
energy needed to raise the temperature of 1 gram of water 1
degree Celsius.
1000 calories = 1 kilocalorie  Calorie in food.
Thermochemistry refers to heat energy. Heat energy moves
between a system under study and its surroundings.
System- What is under study, i.e. contents in a chemical reaction.
Consider a piston containing gases. ├────╨────┤
│H2 O2 N2
│
The gases are the system; piston and everything else are the
surroundings.
Surroundings- Everything other than the system.
Closed System- One in which energy is exchanged with the
surroundings, but not matter.
Force- Push or pull on an object. F = ma, F is force, m is mass in
kilograms, a is acceleration in meters per second square.
Work (w)- Energy used to cause an object to move against a force.
W = F x d, d is distance.
Also, W = -PV ; P is pressure, V is change in volume.
First Law of Thermodynamics- Energy is conserved.
Mathematically, it is expressed this way: U = q + w; U is the internal
energy of the system.
U is the change in internal energy of system. It is a state function.
q- heat energy absorbed (q is positive) or given off (q is negative) by the
system.
w- work done by the system(w is negative) on its surroundings or done
on the system(w is positive) by its surroundings.
Ex. A piston full of gases absorbs 70 kJ of heat, causing the gases in the
piston to expand and do 50 kJ of work on the surroundings. What is the
change in internal energy of the gaseous system?
U = q + w = 70 kJ – 50 kJ = 20 kJ.
Endothermic Process- The change in the system internal energy is
positive (U > 0).
Exothermic Process- The change in the system internal energy is
negative (U <0).
State Function- Value depends only on the initial and final conditions.
U is a state function; so is pressure (P) and volume(V). q and w are not
state functions.
U is associated with a chemical change at constant volume.
H(Enthalpy)- heat energy associated with a chemical reaction at
constant pressure.(i.e. Activity series reactions and Metathesis reactions
done in the lab).
H = U + PV, H is also a state function. Can be endothermic
H = U + (PV)
H = U + (nRT)
(positive) or exothermic (negative).
Associated with all reactions is an enthalpy of reaction, Hrxn
Hrxn = enthalpy for reaction that takes place at standard
pressure(1atm) and temperature(for thermochemistry, it is 298 K, for
ideal gas law, 273 K).
Enthalpy is an extensive property.
Example: CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890kJ/mol
If one doubles the coefficients of this reaction, then the respective
enthalpy change is doubled:
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(l) H = -1780kJ
Enthalpy change for the reverse reaction is equal in magnitude but
opposite in sign(+890kJ).
Enthalpy change for a reaction depends on the state of the reactants and
products.
Example: 2H2O(l)  2H2O(g) H = 88 kJ/mol
2H2O(s)  2H2O(l) H = 12 kJ/mol
Calorimetry- The measurement of heat flow in a reaction.
Calorimeter- An apparatus that measures heat flow
Heat Capacity- The amount of heat energy needed to raise the
temperature of a certain amt. of an object by 1 K or Celcius.
Specific Heat Capacity (Cp)- is amount of heat capacity for 1g of
substance:
Cp = quantity of heat transferred
(grams sub.)(temp. change)
Cp = q/(mT); q = mCpT
Example: 209J is needed to raise the temperature of 50grams of water by
1 K. What is the specific heat?
Therefore Cp = q/(mT) = 209J)/{(50g)(1 K)} = 4.18J/g-K
Heat liberated in a solutionConstant pressure calorimetry: q = (specific heat
soln) x (grams soln) x T
Ex. 50ml 1M HCl soln + 50ml NaOH soln raises temp. of mixture from 21 to
27.5 C.
Assume Cp of soln is 4.18j/g-K; density of soln is 1g/ml.
Then T is 27.5 - 21.0 = 6.5 : total mass is 100ml x 1g/ml = 100g
q = mCp∆T; HCl + NaOH → NaCl + H2O ∆Hrxn = 2.7kJ/0.05mol = 54kJ/mol
q soln. = (4.18J/g-K)(100g)(6.5C) = 2700J (negative because heat was
liberated).
When two masses at different temperatures come into contact, heat from the
warmer mass will transfer to the cooler mass to reach thermal equilibrium (Teq =
equilibrium temperature). Therefore:
qwarm = qcool
Ex. If
20g of warm water (90C) is mixed with 30g of cool water (6.0C), what is
Teq?
mCpT1
=
mCpT2
(20g)(4.184 J/g-K)( 90C-Teq) = (30g)(4.184 J/g-K)(Teq - 6.0C)
7531.2 - 83.68Teq = 125.52Teq - 753.12
209.2Teq = 8284.32
Teq = 39.6C ; Teq is always between warm and cool temperature.
Bomb Calorimerty- For this condition, q = Cpbomb x T ; Cpbomb is heat
capacity of the bomb calorimeter.
Ex. 4.00 grams CH6N2 (Mol Wt. 46 grams) is combusted in a bomb
calorimeter; temp of calorimeter is raised 14.5C; Cp of bomb calorimeter
is 7.794kJ/C. What is heat capacity per mole of compound?
q = 7.794kj/C x 14.5C = 113kJ
4 grams CH6N2 is : (4/46) = 0.087mole , molar heat capacity is
(113kJ/0.087) = 1299kJ/mole
Hess’s Law- If a reaction is carried out in a series of reaction steps, H for
the resultant reaction is equal to the sum of the H’s for the individual step
reactions.
Ex.
Step 1
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) Hrxn1 = -890kJ
Step 2
2H2O(l)  2H2O(g) Hrxn2 = 88kJ
--------------------------------------------------------------------CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = Hrxn1 + Hrxn2 = -802kJ
Standard Enthalpy of Formation (Hf)- For a compound, it is the change in
enthalpy for the reaction that forms 1 mole of the compound from its elements in
their stable states at standard conditions.
For element in their stable states, Hf is zero (i.e. Na atom, O2, H2, S8)
Ex. 2Cgraphite + 3H2(g) + (1/2)O2(g)  C2H5OH
formation of C2H5OH.
Hf = -277.7kJ for the
Then the enthalpy of a reaction at standard (Hrxn) conditions is:
Hrxn = nHf (products) - nHf (reactants)
The Hf values are tabulated in textbooks, usually in kilojoules.
Given
2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l) What is Hrxn?
Hrxn = nHf(products) - nHf(reactants)
Hrxn = [12Hf(CO2) + 6 Hf(H2O)] - [2Hf(C6H6) + 15Hf(O2)]
=
[12(-393.5 kJ) + 6(-285.8 kJ)] - [2(49.0 kJ + 15(0 kJ)]
= -6534.8 kJ for 2 moles of benzene burned OR -3267 kJ/mol
CHAPTER 6 – Thermochemistry Problems & Solutions
Prof. Onwuachi
1. Calculate the total change in internal energy, U, of a
system when 400 J heat is applied to expanding O2 (g) and
the gas does 350 J of work on its surroundings.
ΔU = q + w = 400J – 350J = 50J
2. Calculate the kinetic energy in J, calories, and kJ of a
particle (mass = 9.11x10-28 g) at 6.00x105 m/s.
E = (1/2)mv2 = (1/2)(9.11 x 10-31kg)(6.0 x 105 m/s)2 = 1.64 x 10-19J
= (1.64 x 10-19J) ÷ 4.184J/cal = 3.93 x 10-20 calories
= 1.64 x 10-22kJ
3. Given the following data:
a) 1/2 Br2 (l)  Br (g)
H0 = 111.75 kJ
b) 1/2 Cl2 (g)  Cl (g)
H0 = 121.38 kJ
c) 1/2 Br2 (l) + 1/2 Cl2 (g)  BrCl (g) H0 = 14.7 kJ
determine H0rx for reaction; Br (g) + Cl (g)  BrCl (g)
Reverse a,=> ΔHº = -111.75 kJ
Reverse b, => ΔHº = -121.38 kJ
Add to c. => ΔHº = 14.7 kJ
ΔH = -218.4 kJ
4. Calculate H0 in kJ for reaction;
CuO(s)+ CO (g)  Cu(s) + CO2 (g)
given the following thermochemical data :
(a) 2CO(g) + O2 (g)  2CO2(g)
H0 = - 566.1 kJ
(b) 2Cu(s) + O2 (g)  2CuO(s)
H0 = - 310.5 kJ
Multiply (a) by ½ => ΔHº = -566.1 kJ/2
= -283.05 kJ
Reverse (b) & multiply by ½ => ΔHº = 310.5 kJ/2 = 155.25 kJ
ΔHº
= -128 kJ
5. Consider the following reactions (a) & (b) :
Calculate Hf0 for H2O2 (l) ; H2 (g) + O2 (g)  H2O2 (l)
a) H2O2 (l)  H2O (l) + 1/2 O2 (g) H0 = -98.3 kJ
b) H2 (g) + 1/2 O2 (g)  H2O (l)
H0 = -285.8 kJ
Reverse (a) => ΔHº = 98.3 kJ
Add to (b) => ΔHº = -285.8 kJ
ΔHº = -187.5 kJ
6. A 1.0 g sample of propane, C3H8, was burned in calorimeter.
The temperature rose from 28.5 0C to 32.0 0C and heat of
combustion 10.5 kJ/g. Calculate the heat capacity of the
calorimeter apparatus in kJ/0C .
Equation for bomb calorimeter: q = CbombΔT
10.5 kJ = Cbomb(32.0 ºC – 28.5 ºC)
Cbomb = 3.00 kJ/ºC
7. What is the resulting temperature when 35.0 g of water at 75
0
C is mixed with 15.0 g of water at 10 0C? (Heat capacity (Cp) of
water = 4.184 J/g. 0C)
Equation:
m1CpΔT1 = m2CpΔT2
(35g)(Cp)(75-Tf) = (15g)(Cp)(Tf – 10)
Tf = 55.5 ºC
8. A 20.0 g piece of a metal with specific heat of 0.080 cal/g.0C
at 68 0C dropped into 15.0 ml water in a calorimeter at 23 0C.
Calculate the final equilibrium temperature of the mixture.
Equation: m1CmetalΔT1 = m2CpΔT2
(20g)(0.08cal/g)(68 -Tf) = (15g)(1.0 cal/gºC)(Tf – 23)
Tf = 27.34 ºC
108.8 – 1.6Tf = 15Tf - 345
453.8 = 16.6Tf
453.8/ 16.6 = Tf = 27.34 ºC
9. When 72 g of a metal at 97.00C is added to 100 g of water at
25.0 0C, the final temperature is 45.5 0C. What is the heat capacity
per grams of the metal? (Heat capacity of H2O = 4.184 J/g. 0C)
Equation: m1CmetalΔT1 = m2CpΔT2
(72g)(Cmetal)(97 – 45.5) = (100g)(4.184J/gºC)(45.5 – 25)
Cmetal = 2.31 J/gºC
10. The heat capacity of lead is 0.13 J/g.0C. How many joules of
heat would be required to raise the temperature of 15.0 g of lead
from 22 0C to 38 0C?
Equation : q = mCmetalΔT
= (15g)(0.13J/gºC)(38 – 22)
= 31.2 J
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