Thermochemistry Problems With Answers Page 2
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45.51 The specific heat of iron metal is 0.450 J/g-K. How many J of heat are necessary
to raise the temperature of a 1.05- kg block of iron from 25.0 °C to 88.5 °C?
q = C m ΔT
get C from table or from problem (0.450 J / g-K)
Since C and K have the same "size," the ΔT does not require conversion to K.
ΔT = T
– T
Note: (88.5 – 25.0 = 63.5) gives 1 dp and 3 SF for these #s
f
I
m = 1.05 kg ( 1000 g / 1 kg) = 1050 g
4
q = (0.45 J / g-C) (1050 g ) (88.5 °C – 25.0 °C) = 30003.75 J = 3.00 X 10
J (3 SF)
Answer must show size and SF. Note: Question asked for HOW MANY, so a
positive answer is appropriate.
5.59 What is the connection between Hess’s law and the fact that H is a state function?
ΔH is path-independent so any # of steps will give the true ΔH for the overall
process.
5.61 Calculate the enthalpy change for the reaction
P
O
(s) + 2 O
(g) P
O
(s)
4
6
2
4
10
given the following enthalpies of reaction:
(g) P
EQ 1 P
(s) + 3 O
O
(s)
ΔH = -1640.1 kJ
4
2
4
6
EQ 2 P
(s) + 5 O
(g) P
O
(s)
ΔH = -2940.1 kJ
4
2
4
10
Use Hess's Law. Identify a substance unique to only one DATA equation. Focus on
handling that substance in each case.
REVERSE EQ 1 to get P
O
on left side; reverse sign
4
6
EQ 2 stays same to keep P
O
on right side.
4
10
P
O
(s) P
(s) + 3 O
(g)
ΔH = +1640.1 kJ
4
6
4
2
(g) P
P
(s) + 5 O
O
(s)
ΔH = -2940.1 kJ
4
2
4
10
create "grand" equation and subtract common items
P
O
+ P
+ 5 O
P
+ 3 O
+ P
O
4
6
4
2
4
2
4
10
both P
's leave and, after subtracting 3 O
, get 2 O
's on left side.
4
2
2
P
O
(s) + 2 O
(g) P
O
(s)
4
6
2
4
10
sum the ΔH 's : +1640.1 kJ
-2940.1 kJ
dp rule
ANSWER: ΔH = -1300.0 kJ
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