**Calorimetry**

Design and perform an experiment that will determine the
H_{rxn} for the dissolution of
H_{2}SO_{4} in water.

Allowing for only PV work at constant pressure H = q_{p}; so, we are looking for the amount of heat transferred during the reaction.

How can we determine the heat of a reaction?

When we discussed the transfer of heat to water we measured the temperature change of the water.

Can we measure the temperature change of a reaction? No, we can only measure the temperature of objects. A reaction is not an object; it is a process.

The key to calorimetry is that all of the heat released or absorbed by the thing being studied

is absorbed by the calorimeter and its contents, which are usually water.

So, to find H_{rxn} we
construct a calorimeter, and perform the reaction in the calorimeter.
All the heat released by the reaction is absorbed by the calorimeter.
Since we can measure the change in temperature for the calorimeter we
can determine the amount of heat the calorimeter absorbed.

-q

-q_{rxn} = C T

Since the calorimeter is made of water, a stirrer, a thermometer,
and the container, we break the q_{calorimeter} into two
terms.

This way we do not always have to add exactly

the same amount of water to the calorimeter.

-q_{rxn} = C_{cal} T + m
s T

We can measure the mass of the water which is used, we can measure
the temperature change, and we know the specific heat of water.
However, we do not know C_{cal}!

So, before we can do our experiment we must determine
C_{cal}.

How would you determine C_{cal}? In the equation above
q_{rxn} and C_{cal} are the only unknown quantities.
If q_{rxn} were known, then we could solve for
C_{cal}. We can determine C_{cal} by adding a known
amount of heat and measuring the temperature change.

The easiest way to add a known amount of heat is to add hot water to a calorimeter filled with cold water.

-q_{rxn} = q_{cal} + q_{water}

-m s T_{HW} = C_{cal}
T_{CW} + m s T_{CW}

T for the
calorimeter is the

same as T
for the cold water

Add 50.0 g of water which is at 100.0 °C to our calorimeter which contains 50.0 g of water at 23.0 °C. The final temperature of the calorimeter is 59.0 °C.

-(50.0 g)(4.184 J g^{-1} °C^{-1})(59 °C
- 100 °C) =

C_{cal}(59 °C - 23 °C) + (50.0
g)(4.184 J g^{-1} °C^{-1})(59 °C - 23
°C)

8577.2 J = C_{cal}(36.0) + 7531.2 J

1046 J = C_{cal }(36.0)

C_{cal} = 29.1 J K^{-1}

Now that we know the heat capacity of our calorimeter we can use our calorimeter to determine the amount of heat a reaction releases.

3.20 g of H_{2}SO_{4} were dissolved in 100.0 g of
water, and the temperature went from 23 °C to 29.8 °C.
Determine H for the dissolution of
H_{2}SO_{4} in water.

-q_{rxn} = C_{cal}(29.8-23) + m s
(29.8-23)

Since the H_{2}SO_{4} solution is actually the
thing that absorbs the heat, we have to use the mass of the solution.
Here is where we are going to make a small assumption. The specific
heat capacity of the solution is the same as the specific heat
capacity of water.

-q

q

Since H_{rxn} is expressed in
kJ/mol we are not done yet.

How many moles of acid were dissolved?

So,

summary of calorimetry