Chapter 8
Basic Question 1
8.1.4 Calculate ΔH of a reaction given m combusted, specific heat capacity and the
TG
temperature rise
To change the temperature of a particular calorimeter and the water it contains by one degree
celsius5.0 kJ are needed. The complete combustion of 0.70 gram of ethene gas, C2H4(g), in the
calorimeter causes a temperature rise of 5.3°C. Find the heat of combustion per mole of ethene.
Given : m of ethene = 0.70 g, ΔT = 5.3°C, C = 5.0kJ / ̊ C
R.T.F : ΔH
Q = CΔt = (5.0)(5.3) = 26.5kJ
0.70
n ethane =
= 0.025mole
28
Q 26.5
ΔH = =
= 1060kJ
n 0.025
Basic Question 2
8.1.5 Rewrite equations using ΔH notation per mole of a given reactant or product
_
Given: 3C(s) + 2Fe2O3(s) + 462 kJ → 4Fe(s) + 3CO2(g)
a) Rewrite the equation expressing ΔH as per mole of Fe(s).
3
1
3
C(s) + Fe 2O 3(s) + 115.5kJ → Fe(s) + CO 2(g)
4
2
4
b) Rewrite the equation expressing ΔH as per mole of C(s).
2
4
C(s) + Fe 2O 3(s) + 154kJ → Fe(s) + CO 2(g)
3
3
Basic Question 3
8.1.5 Rewrite equations using ΔH notation per mole of a given reactant or product
_
Given: ½ H2(g) + ½ Br2(l) → HBr(g)
ΔH = -36 kJ/mole HBr
Rewrite the equation expressing the coefficients as the smallest whole numbers, and express the
heat of reaction
a) per mole of H2.
H 2(g) + Br2(g) → 2HBr(g)
ΔH = -72kJ/mole of H 2(g)
8.1.5
Find heat involved with given # of moles of reactant/product from ΔH
_
b) How much energy is liberated when 2 moles of HBr(g) are produced according to the above
equation?
72 kJ are produced when 2 moles of HBr(g) are formed.
Basic Question 4
8.1.5
Find heat involved with given mass of reactant/product from ΔH
_
Consider the following reaction:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH = -819 kJ
Calculate the heat change when 1.60 g of methane is burned in excess oxygen.
Given : mass of CH4 = 1.60 g, ΔH = -819 kJ
R.T.F: heat change (ΔH)due to the given mass of methane
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
16g
1.60g
ΔH= - 81.9 kJ or
ΔH = -819 kJ
819kJ
?kJ
Energy released is 81.9 kJ
Basic Question 5
8.1.6
Application on Hess’s Law
TG
a) In forming water vapour from its elements, 242 kJ/mol are given out. In forming carbon dioxide
from its elements, 393kJ are released. In forming ethane (C2H6) from its elements, 84.5kJ are
produced. Find the heat of combustion of ethane to form water vapour and carbon dioxide.
Given: molar heat of formation of H2O = 242kJ
molar heat of formation of CO2 = 393kJ
molar heat of formation of C2H6 = 84.5kJ
RTF: molar heat of combustion of C2H6
The three equations are:
∆H = - 242 kJ
H2 (g) + ½ O2 (g) → H2O(g)
C + O2 (g) → CO2 (g)
∆H = - 393 kJ
2 C + 3 H2 (g) → C2H6 (g)
∆H = - 84.5 kJ
The equation for the combustion of ethane is:
7
C2H6 (g) + O2 (g) → 2 CO2 (g) + 3 H2O(l)
2
Steps:
1. Multiply equation of formation of water by 3
2. Multiply equation of formation of carbon dioxide by 2
3. Reverse equation for formation of C2H6
3
3 H2 (g) +
O2 (g)→ 3 H2O(g)
∆H = 3 x - 242 kJ
2
∆H = 2 x - 393 kJ
2 C + 2 O2 (g) → 2 CO2 (g)
C2H6 (g)→ 2 C + 3 H2 (g) →
∆H = 84.5 kJ
7
C2 H 6(g) + O2(g) → 2CO2(g) + 3H 2O(g)
2
ΔH = 84.5 + 2(-393) + 3(-242)
ΔH = -1427.5kJ
b) The thermite reaction is a spectacular and highly exothermic reaction. It involves the rection
between Fe2O3 and metallic aluminium. The reaction produces white-hot, molten iron in a few
seconds. Given:
4Al + 3O2→ 2Al2O3
4Fe + 3O2→ 2Fe2O3
ΔH1 = -3348 kJmol
ΔH2 = -1674 kJmol
Determine the amount of heat liberated in the reaction of 1 mole of Fe2O3 with Al.
We need to reverse the second reaction:
2Fe2O3→4Fe + 3O2
4Al + 3O2→ 2Al2O3
ΔH1 = +1674 kJ/mol
ΔH2 = -3348 kJ/mol
Add the two reactions:
ΔH = +1674+ (– 3348) = -1674 kJ/mol
2Fe2O3+ 4Al→ 4Fe + 2Al2O3
We need the reaction per mole of Fe2O3 with Al ⇒ we need to divide the reaction by 2:
Fe2O3 + 2Al →2Fe + Al2O3
ΔH = -837 kJ/mol
Basic Question 6
8.3.3 Be able to identify unknowns in a given a nuclear reaction with missing particles
Fission of uranium gives a variety of fission products, including praseodymium, Pr. In the
equation below:
+
→
+ ____ + 3
+ energy
a) What is the charge on the missing element?
Charge is conserved ⇒ 92 + 0 = 59 + y + 0
⇒
b) What is the mass of the missing element?
Mass is conserved ⇒ 235 + 1 = 147 + x + 3 ⇒
x = 86
c) Identify the element.
As, arsenic
y = 33
T