Enthalpy
Sample Problems:
1. Given the equation 3 O
→2 O
∆H = +285.4 kJ, calculate ∆H for the following
2(g)
3(g)
reaction. 3/2 O
→ O
.
2(g)
3(g)
Answer
Since 3/2 O
→ O
is ½ of 3 O
→2 O
the enthalpy of the reaction will be ½ as
2(g)
3(g)
2(g)
3(g)
well: ½ (+285.4kJ) = +142.7kJ
2. Given the equation: 2Ag
S
+ 2H
O
→ 4Ag
+ 2H
S
+ O
∆H = +595.5kJ,
2
(s)
2
(l)
(s)
2
(g)
2(g)
calculate ∆H for the following reaction.
Ag
+ ½ H
S
+ ¼ O
→ ½ Ag
S
+ ½ H
O
(s)
2
(g)
2(g)
2
(s)
2
(l)
Answer
In this problem, the reaction equation has been reversed and divided by four. The new
enthalpy value will be the opposite sign and ¼ its original value: -1/4(+595.5kJ)
= -148.9kJ
o
3. Express the following information as a chemical equation. At 25
C and at a constant
pressure, dinitrogen trioxide gas decomposes to nitrogen monoxide and nitrogen dioxide
gases with the absorption of 0.533kJ of heat for every gram of dinitrogen trioxide that
decomposes.