Enthalpy Of Reaction Worksheet Page 3
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1
2
3
4Enthalpy Of Reaction (∆
H)
rxn
Model 4: Enthalpy of Reaction using ∆
H
f
In Model 2, you developed a way of working out the value of enthalpy change for a reaction from the values of
enthalpy of atom combination for the reactants and products. From Q7:
Δ
H = Δ
H (products) - Δ
H (reactants)
(4)
rxn
ac
ac
An alternative is to use the enthalpy change of formation of a compound (Δ
H) from its elements in their
f
naturally occurring forms. At room temperature and pressure, these forms are called the standard states of the
elements and include, for example, graphite for carbon and O
(g) for oxygen.
2
Using this method, the equation for the enthalpy of reaction becomes:
Δ
H° = Δ
H° (products) – Δ
H° (reactants)
(5)
rxn
f
f
The enthalpy of formation of CO
(g) is then the energy change for its formation from graphite and O
(g):
2
2
C(s) + ½ O
(g) CO
(g)
2
2
The enthalpy change for the combustion of methane is represented on the energy level diagram below. On the
left, CH
(g) and O
(g) are broken up into their elements in the standard states, graphite (C(s)), H
(g) and O
(g).
4
2
2
2
This is the reverse of their formation so the energy required is –Δ
H° (reactants). On the right, CO
(g) and
f
2
H
O(g) are formed from the same elements in the same states so the energy change is + Δ
H° (products).
2
f
Critical thinking questions
1.
Why are Δ
H° (O
(g)) and Δ
H° (H
(g)) both equal to 0 kJ? (Hint: what is the reaction in each case?)
f
2
f
2
2.
What is Δ
H° for the reaction CH
(g) + 2O
(g) CO
(g) + 2H
O(l)?
rxn
4
2
2
2
3.
Use equation (5) and the data below to calculate Δ
H° for the reaction MgO(s) + CO
(g) MgCO
(s).
rxn
2
3
-1
-1
-1
Δ
H°: MgO(s) = -602 kJ mol
, CO
(g) = -394 kJ mol
and MgCO
(s) = -1096 kJ mol
f
2
3
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