Density massvolume mass densityvolume densityv acid v

density = mass/volume; mass = (density)(volume) = (density)(V
acid
+ V
base
)
m
water
= (1.00 g/mL)(38.5 mL + 42.2 mL) = 80.7 g
C
water
= 4.184 J/g
.o
C
∆T
water
o
C – 25.2
o
C) = 13.4
o
C
From this information we can determine the amount of heat generated.
q
Rxn
= T
f
– T
i
= (38.6
= – (m
water
)(C

water
)(∆T
water
)
= – (80.7 g)(4.184 J/g
.o
C)(13.4
o
C) = – 4.52 x 10
3
J = – 4.52 kJ
This reaction generated –4.52 kJ of energy. Note that the negative sign implies that the reaction is exothermic so a
rise in temperature occurs as the reaction is carried out.
To determine the molar heat of reaction or molar enthalpy of reaction (∆H
Rxn
), the heat of the reaction (q
Rxn
) is divided by the number of moles that reacted, n. Since this reaction is an acid-base neutralization reaction, it
is sometimes referred to as the heat of neutralization (∆H
Neut
).
∆H
Rxn
=
q
Rxn n
We need to determine which reactant is limiting; the one that is completely consumed and controls the amount of
product (or in this case heat) that is generated. Since the stoichiometric relationship between the acid and base is
one-to-one in our example, you can calculate the moles of acid and base:
Moles HNO
3
=
38.5 mL HNO 3
x 1000
1
L mL x
2.2

mol 1
L HNO 3
= 8.5 x 10
–2
mol HNO
3
Moles NaOH =
42.2 mL NaOH x 1000
1
L mL x
1.9
mol 1
L NaOH = 8.0 x 10
–2
mol NaOH
The number of moles of NaOH is slightly smaller that that of HNO
3
, so NaOH is the limiting reactant.
Therefore, ∆H
Rxn
is:
∆H
Rxn
=
q
Rxn n
=
8.0 -
x 4.52 10 2-
kJ mol = – 57 kJ/mol
This means that for each mole of nitric acid that reacts with one mole of sodium hydroxide, 57 kJ will be released.
HEATS OF SOLUTION
When a solute is dissolved in water, heat is absorbed from or released to the surroundings; the enthalpy change for
this process is called the heat of solution or molar enthalpy of solution, ∆H
Soln
. To measure
2

, ∆H
Soln
a solid sample is added to a known amount of water and the temperature change measured. The calculations
are very similar to those used to calculate the ∆H
Rxn
.
Consider the following example: 3.94 g of KBr are added to 50.0 mL of water and the temperature of the water
decreases from 28.5
o
C to 25.4
o
C. Calculate the amount of heat absorbed and the value of ∆H
Soln
.
From the conservation of energy (1
st
Law of Thermodynamics):
q
water
= – q
water
q
Soln
+ q
Soln
= 0 or q
Soln
= – (m)(C
water
)(∆T
water
)
In this case we should remember that both the water and the KBr are undergoing a temperature change, so m in this
case is the mass of the water and the mass of the KBr.
m = m
water
= (50.0 mL x 1.00 g/mL) + 3.94 g = 53.94 g

q
Soln
+ m
KBr
.o
C)(25.4 – 28.5 °C)
q
Soln
= – (53.94 g)(4.184 J/g
= 700 J = 0.70 kJ (note 2 SF because ∆T is to 2 SF)
When 3.94 g of KBr (MM = 119.00 g/mol) are dissolved it consumes 0.77 kJ of energy since the process is
endothermic. To determine how much energy is consumed per mole simply divide the heat by the number of moles
of solute that dissolved:
∆H
Soln
=
q
Soln n
= 3.31
0.70
x 10 2-
kJ mol = 21 kJ/mol
The experimentally-determined heat of solution for potassium bromide is 21 kJ/mol. The accepted value (published
in the chemical literature) for the heat of solution of KBr is 19.87 kJ/mol. The percent error is 6% which is