Calorimetry booklet (October 2014): Q8)
Info in question given: HNO3(aq) + Ba(OH)2(aq)
Acid + base --> salt + water. A salt is a compound containing a cation and an anion. Barium,
is a Group II metal, and like all group II metals, is a +2 ion in compounds. When acting as an
acid, nitric acid, H-NO3, leaves behind nitrate ions, NO3-, after reaction. Nitrate anions have a
-1 charge, so two nitrate ions are needed to balance the charge of the Ba2+ ion. Water will
be produced from the -OH ion in Ba(OH)2, and H from the H-NO3 acid.
HNO3(aq) + Ba(OH)2(aq) --> Ba2+(aq) + 2 NO3- (aq) + H2O (l)
We need two H2O's to use up the two OH's in Ba(OH)2
HNO3(aq + Ba(OH)2(aq) --> Ba2+(aq) + 2 NO3- (aq) + 2 H2O (l)
so we need two HNO3 to supply both the remaining two H in 2 H2O and to give the right
number of nitrate anions.
2HNO3(aq) + Ba(OH)2(aq) --> Ba(NO3)2(aq) + 2 H2O(l)
^^^ The balanced chemical equation (in whole number ratios) ^^^
Formula's used (MEMORISE THESE)
'worded
equation'
Concentration =
moles present /
volume in L
Symbol form
Units
Notes
c=n/V
Mol dm-3
Volume MUST be in litres. A litre
(symbol 'L') is the same as 1 dm3.
There are 1000 cm3 in 1L and there are
1000 L in a cubic metre.
In z cm3 there are (z/1000) L
moles present =
mass used / mass
of one mole.
n = m / Mr (for
molecules)
n = m / Ar (for
atomic species)
H = mcT
Enthalpy change
= (mass) x
(specific heat) x
temperature
change.
Sometimes
expressed as
H = mc
mass of substance whose temperature
increase is to be measured.
specific heat capacity of substance
whose temperature is to be changed
Usually, formula's are constructed to
use Kelvin values of temperatures, but
when we have changes in temp, we
can use Celsius (Centigrade) or Kelvin it
doesn't matter here.
Yield limiting reagent:
The balanced chemical equation tells us the relative amounts of substances needed to
achieve the reaction and get the maximum amount of product. However we may not
actually have the necessary amount of one of the reactants, hence this will limit the amount
of product that can form and is called the yield limiting reagent.
e.g. 20 people may be necessary to build 4 homes in 6 months. If only 10 people are present
then we cannot build 4 homes in 6 months, but a smaller more limited number instead.
We check to see which is the yield limiting reagent by calculating the moles of each
substance and then comparing it to the balanced chemical equation.
moles of HNO3 present = c * V ⇨ n = 1.0 * (50/1000) = 0.050 moles (keep s.f.)
moles of Ba(OH)2 present = c * V ⇨ n = 1.0 * (20/1000) = 0.020 moles (keep s.f.)
The balanced chemical equation tells us that for every one Ba(OH)2 we have, then 2
HNO3 are needed. From our calculations there are more than .04 moles of HNO3 present
to cope with the 0.020 moles of Ba(OH)2 present. It is actually the Ba(OH)2 that is the
yield limiting reagent, as there isn't enough barium hydroxide to react with all the HNO3
So, bearing in mind the ratio's in the balanced chemical equation. what's actually
happening in our reaction is:
0.04 HNO3(aq) + 0.02 Ba(OH)2(aq) --> 0.02 Ba(NO3)2(aq) + 0.04 H2O(l)
^^^ The enthalpy we calculate will correspond to THIS equation ^^^
The question asks us to scale this up per mole of HNO3, so to transform 0.04 into 1, we
multiply by 25, and must multiply all other quantities by 25 also to maintain balance.
1 HNO3(aq) + 0.5 Ba(OH)2(aq) --> 0.5 Ba(NO3)2(aq) + 1 H2O(l)
Hence at the end we will scale up the enthalpy value so that it represents the equation for
one mole of HNO3. BEWARE: THIS 'scale-up step' is sometimes forgotten by students.
Temperature change:
In this case we are given the T (delta T) value to use in the equation H = mcT but
sometimes we must calculate it ourselves. The correct way to do this (or any  term of a
chemical reaction) is as follows:
FINAL state minus INITIAL state
(in some other instances expressed as: "products minus reactants")
We are told about the reaction given, that the temperature of the water rose by(increased)
7.9oC. The process of water increasing in temperature (corresponding to an increase in
enthalpy) is actually an endothermic process, T(final) - T(initial) is positive. Putting the
positive T and the 'm' and 'c' values (which are also positive) in the H = mcT equation
will give a positive H value. Yet in this case, the correct answer would be a negative value
for H.
What we are actually doing in this and most other A-level calorimetry experiments, is
measuring the enthalpy change of the aqueous system, and NOT of the chemical reaction.
But we use the enthalpy change of the aqueous system to (indirectly) deduce what
happened in the chemical system.
This is perhaps well illustrated by the more accurately described calculation steps:
H(aqueous system) = m(aqueous system) x c(aqueous system) x T(aqueous system)
and then state
H(aq system) = -H(chemical reaction)
Because what happens in the water is the exact opposite if the chemical system, i.e. If the
chemical system gains enthalpy, that energy would have had to have come from the
surrounding water, and hence the water would have lost enthalpy which would have caused
a reduction in temperature in the water. Conversely, as is happens to be the case here, If
the chemical system loses energy enthalpy, it would be given off to the surrounding water
and hence, the water system would have gained enthalpy, causing an increase in the
temperature in the water.
But nobody ever does this. Shortcuts are employed, which assume nobody will ever start
thinking about what they are actually doing or what actually happens. But if you do begin to
start thinking about it, you may feel worried of the un-mathematical nature of just pulling a
minus out of thin air, for endothermic reactions just throwing away an minus.
Ok to the calculation.
H = mcT
volume of water = 70cm3. Density of water = 1 g for every cm3 . There are 70 cm3 so there
are 70g of water.
m = 70 g, c = 4.18 J g-1 K-1 , T = 7.9, H = 70 x 4.18 x 7.9 = +2311.54
Always write a PLUS or a MINUS when you give a H value!
Remember our previous discussion? -: the enthalpy change corresponds to the enthalpy
change of the water (the aqueous environment). The actual chemical reaction has the
opposite sign!
H(water)= mcT = -H(chemical reaction)
Hence = H(chemical reaction) = - 2311.54
Units? (Always consider units after a calculation, way too many student lose marks for not
stating the units after a calculation, even though it's an almost certain case that they
students did actually know what the unite were!)
The units of H are kJ mol-1, but we don't have a mole of anything in the reaction that that
we did. This reveals a strange thing about H, that the units are always kJ mol-1 even when
we don't have one mole present. Thus to stop any possible confusion:
H values must always be accompanied by a chemical equation.
I.e. the H term is PER THE MOLES STATED IN THE GIVEN EQUATION.
Hence for
0.04 HNO3(aq) + 0.02 Ba(OH)2(aq) --> 0.02 Ba(NO3)2(aq) + 0.04 H2O(l)
H = -2.31154 kJ mol-1
The question asks us to calculate H per mole of HNO3 ie. the scale-up step we mentioned
earlier (we multiply everything by 25)
1 HNO3(aq) + 0.5 Ba(OH)2(aq) --> 0.5 Ba(NO3)2(aq) + 1 H2O(l)
H = - 57.7885 kJ mol-1
And to give the correct number of s.f's (two s.f's in this case) we round up and report
H = - 58 kJ mol-1