CHEM 103 ORGANIC CHEMISTRY PORTAGE LEARNING MODULE 3
3.2: THERMOCHEMICAL EQUATIONS & MEASURING HEATS OF REACTIONS
The equation written below is a thermochemical equation because it shows that
three moles of H2 gas react with 1 mole of N2 gas to form two moles of NH3 gas
and 91.8 kJ (kilojoules) of heat is given off.
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔHrx = -91.8 kJ
Two important rules that apply to thermochemical equations are:
(1) When the reverse of the thermochemical equation occurs, the sign of the ΔH is
reversed
(2) When the number of moles of reactants used is changed, the quantity of heat
absorbed or evolved is equal to the original value of ΔH times the factor (new
moles/original moles).
moles refers to substance for which mass data is given in the problem
q = ΔHrx x new moles (moles given in problem) / original moles (moles given in
thermochemical equation)
The heat involved in the reaction of 51 grams NH3 to produce N2 and H2 is
calculated below (note that this is the reverse of the original reaction):
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔHrx = -91.8 kJ
2 NH3 (g) → 3 H2 (g) + N2 (g)
ΔHrx is for 2 mole of NH3
reaction uses 51 g NH3 = 51/17 mole = 3 mole NH3
q = + 91.8 kJ x 3 mole NH mole NH3 = + 137.7 kJ
Measuring Heats of Reactions
To be able to determine how much heat is given off (or absorbed) in these two
reactions (C + O2 → CO2 and HCl + NaOH → NaCl + H2O), we must be able to
determine the amount of heat absorbed by the water as its temperature
increases. This quantity of heat is determined by multiplying three pieces of
information about the water: specific heat, temp change, and mass.
,The heat (given off by the reaction) and absorbed by the water in the
reaction of 200 ml of 1.0 M NaOH and 200 ml of 1.0 M HCl causes the
temperature of the water and the calorimeter (heat capacity of calorimeter
= 340 J/oK) to increase by 5.0oK (note total mass of water is 400 grams
since the water in the two solutions (HCl and NaOH) is 400 ml = 400
grams).
HCl + NaOH → NaCl + H2O
c (water) = 4.184 J/g oC
q water = s (specific heat of water) x mass x Δt = 4.18 J / g / oK x 400 g x 5.0oK
= -8360 J
q calorimeter = heat capacity x Δt = 340 J/oK x 5.0oK = -1700 J
(signs are negative since heat is given off, exothermic)
q reaction = -1700 J + (-8360 J) = -10060 J = -10060 J x 1 kJ / 1000 J = -10.06
kJ
new moles of HCl or NaOH = (1.00 M/L) x (0.200 L) = 0.200 mole
ΔH reaction = -10.06 kJ / (0.200 mol / 1 mole) = -50.3 kJ / mole
1 HCl + 1 NaOH → 1 NaCl + 1 H2O
ΔH = -50.3 kJ / mole
The heat (given off) by the combustion of graphite and absorbed by 2000 g
of water (in the reaction of 1.00 g of C (graphite) and excess O2 causes the
temperature of the water and the calorimeter (heat capacity of calorimeter
= 21.0 kJ/oK) to increase by 1.6oK
graphite; a gray crystalline allotropic form of carbon
C + O2 → CO2
c (water) = 4.184 J/g oC
q water = s x mass x Δt = 4.18 J / g / oK x 2000 g x 1.6oK = -13376 J
q calorimeter = heat capacity x Δt = 21.0 kJ / oK x 1.6oK = -33.6 kJ x 1000 J/1
kJ= - 33600 J
q reaction = -33600 J + (-13376 J) = -46976 J = -46976 J x 1 kJ / 1000 J = -
46.976 kJ
, new moles = 1.00 g / 12 = 0.08333 mole
ΔH reaction = -46.976 kJ / (0.) = -563.7 kJ / mole
C + O2 → CO2
ΔH = -563.7 kJ / mole
Thermochemical Equation Problems
1. Ammonia undergoes combustion to yield nitric oxide and water by the
following reaction equation:
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) ΔH = - 1170 kJ
If 26.5 g of NH3 is reacted with excess O2, what will be the amount of heat
given off?
ΔHrx is for 4 mole of NH3
reaction uses 26.5 g NH3 = 26.5/17 = 1.56 mole NH3
q = -1170 kJ x 1.56 mole NH mole NH3) = - 456.3 kJ
Thermochemical Equation Problems
2. Sulfur undergoes combustion to yield sulfur trioxide by the following
reaction equation:
2 S + 3 O2 → 2 SO3 ΔH = - 792 kJ
If 42.8 g of S is reacted with excess O2, what will be the amount of heat
given off?
ΔHrx is for 2 mole of S
reaction uses 42.8 g S = 42.8/32.06 = 1.335 mole S
q = -792 kJ x 1.335 mole S / 2 mole S = -528.7 kJ
Thermochemical Equation Problems
3. Methane (CH4) reacts with Cl2 to yield CCl4 and HCl by the following
reaction equation:
CH4 + 4 Cl2 → CCl4 + 4 HCl
What is the ΔH of the reaction if 51.3 g of CH4 reacts with excess Cl2 to yield
1387.6 kJ?
ΔHrx is for 1 mole of CH4; q = -1387.6kJ
reaction uses 51.3 g CH4 = 51.3/16.042 mole = 3.198 mole CH4
3.2: THERMOCHEMICAL EQUATIONS & MEASURING HEATS OF REACTIONS
The equation written below is a thermochemical equation because it shows that
three moles of H2 gas react with 1 mole of N2 gas to form two moles of NH3 gas
and 91.8 kJ (kilojoules) of heat is given off.
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔHrx = -91.8 kJ
Two important rules that apply to thermochemical equations are:
(1) When the reverse of the thermochemical equation occurs, the sign of the ΔH is
reversed
(2) When the number of moles of reactants used is changed, the quantity of heat
absorbed or evolved is equal to the original value of ΔH times the factor (new
moles/original moles).
moles refers to substance for which mass data is given in the problem
q = ΔHrx x new moles (moles given in problem) / original moles (moles given in
thermochemical equation)
The heat involved in the reaction of 51 grams NH3 to produce N2 and H2 is
calculated below (note that this is the reverse of the original reaction):
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔHrx = -91.8 kJ
2 NH3 (g) → 3 H2 (g) + N2 (g)
ΔHrx is for 2 mole of NH3
reaction uses 51 g NH3 = 51/17 mole = 3 mole NH3
q = + 91.8 kJ x 3 mole NH mole NH3 = + 137.7 kJ
Measuring Heats of Reactions
To be able to determine how much heat is given off (or absorbed) in these two
reactions (C + O2 → CO2 and HCl + NaOH → NaCl + H2O), we must be able to
determine the amount of heat absorbed by the water as its temperature
increases. This quantity of heat is determined by multiplying three pieces of
information about the water: specific heat, temp change, and mass.
,The heat (given off by the reaction) and absorbed by the water in the
reaction of 200 ml of 1.0 M NaOH and 200 ml of 1.0 M HCl causes the
temperature of the water and the calorimeter (heat capacity of calorimeter
= 340 J/oK) to increase by 5.0oK (note total mass of water is 400 grams
since the water in the two solutions (HCl and NaOH) is 400 ml = 400
grams).
HCl + NaOH → NaCl + H2O
c (water) = 4.184 J/g oC
q water = s (specific heat of water) x mass x Δt = 4.18 J / g / oK x 400 g x 5.0oK
= -8360 J
q calorimeter = heat capacity x Δt = 340 J/oK x 5.0oK = -1700 J
(signs are negative since heat is given off, exothermic)
q reaction = -1700 J + (-8360 J) = -10060 J = -10060 J x 1 kJ / 1000 J = -10.06
kJ
new moles of HCl or NaOH = (1.00 M/L) x (0.200 L) = 0.200 mole
ΔH reaction = -10.06 kJ / (0.200 mol / 1 mole) = -50.3 kJ / mole
1 HCl + 1 NaOH → 1 NaCl + 1 H2O
ΔH = -50.3 kJ / mole
The heat (given off) by the combustion of graphite and absorbed by 2000 g
of water (in the reaction of 1.00 g of C (graphite) and excess O2 causes the
temperature of the water and the calorimeter (heat capacity of calorimeter
= 21.0 kJ/oK) to increase by 1.6oK
graphite; a gray crystalline allotropic form of carbon
C + O2 → CO2
c (water) = 4.184 J/g oC
q water = s x mass x Δt = 4.18 J / g / oK x 2000 g x 1.6oK = -13376 J
q calorimeter = heat capacity x Δt = 21.0 kJ / oK x 1.6oK = -33.6 kJ x 1000 J/1
kJ= - 33600 J
q reaction = -33600 J + (-13376 J) = -46976 J = -46976 J x 1 kJ / 1000 J = -
46.976 kJ
, new moles = 1.00 g / 12 = 0.08333 mole
ΔH reaction = -46.976 kJ / (0.) = -563.7 kJ / mole
C + O2 → CO2
ΔH = -563.7 kJ / mole
Thermochemical Equation Problems
1. Ammonia undergoes combustion to yield nitric oxide and water by the
following reaction equation:
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) ΔH = - 1170 kJ
If 26.5 g of NH3 is reacted with excess O2, what will be the amount of heat
given off?
ΔHrx is for 4 mole of NH3
reaction uses 26.5 g NH3 = 26.5/17 = 1.56 mole NH3
q = -1170 kJ x 1.56 mole NH mole NH3) = - 456.3 kJ
Thermochemical Equation Problems
2. Sulfur undergoes combustion to yield sulfur trioxide by the following
reaction equation:
2 S + 3 O2 → 2 SO3 ΔH = - 792 kJ
If 42.8 g of S is reacted with excess O2, what will be the amount of heat
given off?
ΔHrx is for 2 mole of S
reaction uses 42.8 g S = 42.8/32.06 = 1.335 mole S
q = -792 kJ x 1.335 mole S / 2 mole S = -528.7 kJ
Thermochemical Equation Problems
3. Methane (CH4) reacts with Cl2 to yield CCl4 and HCl by the following
reaction equation:
CH4 + 4 Cl2 → CCl4 + 4 HCl
What is the ΔH of the reaction if 51.3 g of CH4 reacts with excess Cl2 to yield
1387.6 kJ?
ΔHrx is for 1 mole of CH4; q = -1387.6kJ
reaction uses 51.3 g CH4 = 51.3/16.042 mole = 3.198 mole CH4
