Chemistry 11 (HL)
Unit 5 / IB Topic 5.1
Energetics 1 Practice Problems
5.1 Exothermic and Endothermic Reactions
Reference: Higher Level Chemistry, p. 158 - 163
1.
Classify these reactions as exothermic or endothermic:
a) 2H2(g) + O2(g) → 2H2O(l)
2.
∆H = - 285.6 kJ
exo
b) Sn(s) + Cl2(g) → SnCl2(s) + 325.1 kJ
exo
c) H2O(l) + 6.0 kJ → H2O(g)
endo
d)
N2(g) + 2H2O(g) → N2H4(l) + O2(g)
e)
C(s) + O2(g) + 393.5 kJ → CO2(g)
∆H = 534.2 kJ
endo
endo
When magnesium is added to dilute sulfuric acid, the temperature of the acid rises.
a)
Is the reaction endothermic or exothermic? exothermic
b)
Draw an enthalpy diagram for this reaction.
Mg + H2SO4
H
∆H
H2 + MgSO4
reaction progress
c)
Write a balanced chemical equation for this reaction.
Mg(s) + H2SO4(aq)  H2(g) + MgSO4(aq)
3.
d)
What is the sign of the enthalpy change for this reaction?
– ∆H
e)
Explain what the enthalpy change implies about the chemical potential
energy of the reactants and products.
Consider “chemical potential energy” as enthalpy. A negative ∆H implies that
the enthalpy of the products is LESS than the enthalpy of the reactants.
Which statements about exothermic reactions are correct?
I. They have negative ∆H values. √
II. The products have a lower enthalpy than the reactants. √
III. The products are more energetically stable than the reactants. √
(lower energy = greater stablility)
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
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Chemistry 11 (HL)
4.
Unit 5 / IB Topic 5.1
When the solids Ba(OH)2 and NH4SCN are mixed, a solution is produced and the
temperature drops.
Ba(OH)2(s) + 2NH4SCN(s) → Ba(SCN)2(aq) + 2NH3(g) + 2H2O(l)
Which statement about the energetics of this reaction is correct?
A.
The reaction is endothermic and ∆H is negative.
B.
The reaction is endothermic and ∆H is positive.
C.
The reaction is exothermic and ∆H is negative.
D.
The reaction is exothermic and ∆H is positive.
5.
Given this enthalpy diagram:
Which statement is CORRECT?
HINT: Substances are more stable at low energy.
A.
B.
C.
D.
Stability of reactants and products
Products are more stable than reactants. x
Products are more stable than reactants. x
Reactants are more stable than products. √
Reactants are more stable than products. √
Type of reaction
endothermic √
exothermic x
endothermic √
exothermic x
6. The combustion of 1 mole of liquid benzene (C6H6) in oxygen liberates 3.268 x 103 kJ of
heat. The products of the reaction are carbon dioxide and water.
a) Is the reaction exothermic or endothermic?
exothermic (energy liberated = released)
b) Write the thermochemical equation for this reaction.
C6H6(l) + 15/2 O2(g)  6 CO2(g) + 3 H2O(l)
3
∆H = -3.268 x 10 kJ
–1
(or kJ mol )
c) Draw an enthalpy diagram for this reaction.
C6H6 + 15/2 O2
H
∆H
CO2 + H2O
reaction progress
d) Calculate the amount of heat evolved when 10.0 g of benzene are burned.
moles C6H6 = 10.0 g x
heat = 0.128 mol x
1 mol
= 0.128 mol
78.12 g
3.268 x 103 kJ
= 418 kJ
1 mol
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Chemistry 11 (HL)
Unit 5 / IB Topic 5.1
e) What mass of benzene must be burned to release 2.0 MJ of heat?
moles C6H6 = 2.0 x 10 3 kJ x
mass C6H6 = 0.612 mol x
1 mol
= 0.612 mol
3.268 x 103 kJ
78.12 g
= 48 g
1 mol
7. The decomposition of 146 g of gaseous HCl into hydrogen and chlorine gases requires
370 kJ of heat.
a)
Write the thermochemical equation for this reaction. (Remember to calculate the ∆H
based on the amount of reactant given in the balanced equation.)
Step 1:
2 HCl(g)  H2(g) + Cl2(g)
Steps 2 and 3:
moles HCl = 146 g x
1 mol
= 4.01 mol
36.45 g
heat required per mole HCl =
370 kJ
= 92.3 kJ mol –1
4.01 mol
heat for 2 moles HCl = (92.3 kJ mol−1 ) x 2 mol = 184.6 kJ
∴ ∆H = +185 kJ
Step 4: 2 HCl(g)  H2(g) + Cl2(g)
∆H = +185 kJ
b) Calculate the amount of heat required for the decomposition of 1.00 kg of HCl.
8.
moles HCl = 1.00 x 103 g x
1 mol
= 27.4 mol
36.45 g
heat required = 27.4 mol x
92.3 kJ
= 2.53 x 103 kJ
1 mol
The enthalpy change for the vaporization of methanol (CH3OH) at 298 K is 37.43
kJ/mol. What quantity of heat energy must be used to vaporize 25.0 g of methanol at
this temperature?
moles CH 3 OH = 25.0 g x
1 mol
= 0.780 mol
32.05 g
heat required = 0.780 mol x
37.43 kJ
= 29.2 kJ
1 mol
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Chemistry 11 (HL)
Unit 5 / IB Topic 5.1
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