Calorimetry Practice Worksheet (Section 17.2)
1.) Compound A is burned in a bomb calorimeter that contains 2.50 liters of water. If the
combustion of 0.175 moles of this compound causes the temperature of the water to
rise 45.00C, what is the molar heat of combustion of compound A (in kJ/mole)? The
heat capacity of water is 4.18 J/g 0C.
∆H = -q
nH = mc ∆t
(0.175 moles) H = (2500 g)(4.18 J/g0C)(45.00C)
H = (2500 g)(4.18 J/g0C)(45.00C)
0.175 moles
H = - 2687142.857 J/mole = -2.69 x 103 kJ/mole
2.) Compound B is burned in a bomb calorimeter that contains 1.50 liters of water. When I
burned 50.0 grams of compound B in the calorimeter, the temperature rise of the water
in the calorimeter was 35.00C. If the heat of combustion of compound B is 2150
kJ/mole, what is the molar mass of compound B? The heat capacity of water is
4.18 J/g 0C.
∆H = q
nH = mc ∆t
n (2150 kJ/mole) = (1500 g)(4.18 J/g0C)(35.00C)
n = (1500 g)(4.18 J/g0C)(35.00C)
2.150 x 106 J/mole
n = 0.102 moles
M = m = 50.0 g =
490 g/mole
n 0.102 moles
3.) The molar heat of combustion of compound C is 1250 kJ/mol. If I were to burn 0.115
moles of this compound in a bomb calorimeter with a reservoir that holds 2.50 liters of
water, what would the expected temperature increase be?
∆H = q
nH = mc ∆t
(0.115 moles)(1250 kJ/mole) = (2500 g)(4.18 J/g0C)(∆t)
∆t = (0.115 moles)(1.250 x 106 J/mole)
=
13.8 0C
(2500 g)(4.18 J/g0C)
4.) A calorimeter was calibrated with an electric heater, which supplied 22.5 kJ of energy to
the calorimeter and increased the temperature of the calorimeter and its water from
22.450C to 23.970C. What is the heat capacity of the calorimeter?
∆H = q
∆H = C ∆t
22.5 kJ = C (22.450C - 23.970C)
C = 22.5 kJ
=
14.8 kJ/0C
1.520C
5.) When 0.113 g of benzene, C6H6, burns in excess oxygen in a calibrated constantpressure calorimeter with a heat capacity of 551 J/0C, the temperature of the
calorimeter rises by 8.600C. Write the thermochemical equation and calculate the
reaction enthalpy for the combustion of benzene (which yields carbon dioxide vapour
and liquid water).
∆H = q
nH = C ∆t
(0.113 g)
H
=
551 J/0C (8.600C)
78.11 g/mole
H = (551 J/0C) (8.600C)
=
- 3.27 x 103 kJ/mole
(0.00145 moles)
C6H6(l) +
C6H6(l) +
15/2 O2(g)
15/2 O2(g)
6CO2(g)
6CO2(g)
+
+
3H2O(l)
3H2O(l);
+
3270 kJ
∆H = -3270 kJ
Multiple Choice:
1.) ____ The value of ΔHo for the reaction below is +128.l kJ:
CH3OH (l)
CO (g) + 2H2 (g)
How many kJ of heat are consumed when 5.10 g of H2 (g) is formed as shown in
the equation? Solution: 5.10 g 1 mole H2 =2.52 moles H2
2.02 g
From the equation: 128.1 kJ/2 moles of H2 = 64.05 kJ/mole H2
So, ∆H = nH
∆H = (2.52 moles H2)(64.05 kJ/mole)
(a) 653
(b) 62.0
(c) 324
(d) 162
2.) ____ The value of ΔHo for the reaction below is +128.1 kJ:
CH3OH (l)
CO (g) + 2H2 (g)
How many kJ of heat are consumed when 5.10 g of CO (g) is formed as shown in the
equation? Solution: 5.10 g
1 mole CO =0.182 moles CO
28.01 g
From the equation: 128.1 kJ/1 moles of CO = 128.1 kJ/mole CO
So, ∆H = nH
∆H = (0.182 moles CO)(128.1 kJ/mole)
(a)
0.182 kJ
(b) 162 kJ
(c) 8.31 kJ
(d) 23.3 kJ
3.) ____ The value of ΔHo for the reaction below is -1107 kJ:
2Ba (s) + O2 (g)
Solution: 5.75 g
2 BaO (s)
1 mole BaO= 0.0375 moles BaO
153.33
g
From the equation: -1107 kJ/2 moles of BaO = - 553.5 kJ/mole BaO
So, ∆H = nH
∆H = (0.0375 moles BaO)(553.5 kJ/mole)
How many kJ of heat are released when 5.75 g of BaO (s) is produced?
(a) 56.9 kJ
(b) 23.2 kJ
(c) 20.8 kJ
(d) 193 kJ
4.) ____ A sample of aluminum metal absorbs 9.86 J of heat, upon which the
temperature of the sample increased from 23.2oC to 30.5oC. Since the specific
heat capacity of aluminum is 0.90 J/g-K, the mass of the sample is
__________g. Note: 23.2oC = 296.2 K and 30.5oC = 303.5 K
q = mc ∆t
9.86 J = (m)(0.90 J/g K)( 7.3 K)
m = 9.86 J
(0.90 J/g K)( 7.3 K)
(a)
72
(b) 65
(c) 1.50
(d) 8.1