SME
General Chemistry I · Problem type
ΔU and ΔH from a bomb calorimeter
Use this for “0.6400 g of naphthalene, C₁₀H₈ (128.17 g/mol), burns in a bomb calorimeter of 10.00 kJ/°C, and the water warms by 2.570 °C. Find ΔU and ΔH of combustion at 25 °C.”
Diagram
q = calorimeter constant × ΔT = 10 × 2.57 = 25.7 kJ
The
volume
is
fixed,
so
no
work
is
done
and
the
heat
is
ΔU:
ΔU
=
−
q
n
=
−
25.7
0.004993
=
−5,147
kJ/mol
of
naphthalene.
ΔH = ΔU + Δn(gas)RT
ΔH = −5,147 + (−2) × 0.008314 × 298.15
ΔH = −5,152 kJ/mol
This is an example. Type your own values to start a new problem.
m
Mass of the sample
example · g
M
Molar mass of the sample
example · g/mol
n
Moles burned
calculated · mol
Ccal
Calorimeter constant
example · kJ/°C
ΔT
Temperature rise
example · °C
q
Heat the calorimeter takes in
calculated · kJ
ΔU
Energy change of combustion per mole
calculated · kJ/mol
Δng
Change in moles of gas
example
T
Temperature
example · K
ΔH
Enthalpy of combustion per mole
calculated · kJ/mol
Formulas
n
=
m
M
0.004993
=
0.64
128.17
q
=
Ccal
×
ΔT
25.7 = 10 × 2.57
ΔU
=
−
q
n
−5,147
=
−
25.7
0.004993
ΔH
=
ΔU
+
Δng
×
0.008314
×
T
−5,152 = −5,147 + (−2) × 0.008314 × 298.15
Assumptions
•
The bomb is sealed, so the volume stays constant and no work is done: the heat the calorimeter takes in is −ΔU of the burning.
•
The calorimeter constant Ccal covers the bomb, the bucket and its water together, and no heat leaks out of the jacket.
•
Δng is moles of gas made less moles of gas used, from the balanced equation with water as a liquid: C₁₀H₈ + 12O₂ → 10CO₂ + 4H₂O gives 10 − 12 = −2.
•
R = 0.008314 kJ/(mol·K) and T is in kelvins (25 °C is 298.15 K).
Step-by-step
WE KNOW
m = 0.64 g, M = 128.17 g/mol, Ccal = 10 kJ/°C, ΔT = 2.57 °C, Δng = −2, T = 298.15 K
FIND
moles burned (n), heat the calorimeter takes in (q), energy change of combustion per mole (ΔU), enthalpy of combustion per mole (ΔH)
Step 1 · Find moles burned (n)
Use n = m ÷ M
Grams over the molar mass counts the moles burned.
n
=
0.64
128.17
n
=
0.004993
mol
Step 2 · Find heat the calorimeter takes in (q)
Use q = Ccal × ΔT
The whole calorimeter takes in Ccal kilojoules for each degree it warms.
q
=
10
×
2.57
q
=
25.7
kJ
Step 3 · Find energy change of combustion per mole (ΔU)
Use ΔU = −q ÷ n
The
burning
sample
gave
off
what
the
calorimeter
took
in,
so
its
heat
is
−
q
.
At
constant
volume
that
is
ΔU;
share
it
among
the
moles.
ΔU
=
−
25.7
0.004993
ΔU = −5,147 kJ/mol
Step 4 · Find enthalpy of combustion per mole (ΔH)
Use ΔH = ΔU + Δng × 0.008314 × T
At constant pressure the gases made or used would push on the air: add Δng RT, with R in kJ.
ΔH = −5,147 + (−2) × 0.008314 × 298.15
ΔH = −5,147 + (−0.01663) × 298.15
ΔH = −5,147 + (−4.958)
ΔH = −5,152 kJ/mol
Check
Put the numbers back in. Both sides should match.
0.004993
=
0.64
128.17
✓
25.7 = 10 × 2.57 ✓
−5,147
=
−
25.7
0.004993
✓
−5,152 = −5,147 + (−2) × 0.008314 × 298.15 ✓
Topic check
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