Constant-Pressure Calorimetry

Constant-Pressure Calorimetry

A constant-pressure calorimeter measures the change in enthalpy of a reaction occurring in a liquid solution. In that case, the gaseous pressure above the solution remains constant, and we say that the reaction is occurring under conditions of constant pressure. The heat transferred to/from the solution in order for the reaction to occur is equal to the change in enthalpy ($\Delta H = q_P$), and a constant-pressure calorimeter thus measures this heat of reaction. In contrast, a bomb calorimeter's volume is constant, so there is no pressure-volume work and the heat measured relates to the change in internal energy ($\Delta U=q_V$).

A simple example of a constant-pressure calorimeter is a coffee-cup calorimeter, which is constructed from two nested Styrofoam cups and a lid with two holes, which allows for the insertion of a thermometer and a stirring rod. The inner cup holds a known amount of a liquid, usually water, that absorbs the heat from the reaction. The outer cup is assumed to be perfectly adiabatic, meaning that it does not absorb any heat whatsoever. As such, the outer cup is assumed to be a perfect insulator.

Coffee cup calorimeter

A styrofoam cup with an inserted thermometer can be used as a calorimeter, in order to measure the change in enthalpy/heat of reaction at constant pressure.

Calculating Specific Heat

Data collected during a constant-pressure calorimetry experiment can be used to calculate the heat capacity of an unknown substance. We already know our equation relating heat (q), specific heat capacity (C), and the change in observed temperature ($\Delta T$) :

$q=mC\Delta T$

We will now illustrate how to use this equation to calculate the specific heat capacity of a substance.

Example 1

A student heats a 5.0 g sample of an unknown metal to a temperature of 207 $^\circ$C, and then drops the sample into a coffee-cup calorimeter containing 36.0 g of water at 25.0 $^\circ$C. After thermal equilibrium has been established, the final temperature of the water in the calorimeter is 26.0$^\circ$C. What is the specific heat of the unknown metal? (The specific heat of water is 4.18 $\frac {J} {g^\circ C}$)

The walls of the coffee-cup calorimeter are assumed to be perfectly adiabatic, so we can assume that all of the heat from the metal was transferred to the water:

$-q_{metal}=q_{water}$

Substituting in our above equation, we get:

$-m_{metal}C_{metal} \Delta T_{metal}=m_{water}C_{water}\Delta T_{water}$

Then we can plug in our known values:

$-(5.0\;g)C_{metal}(26.0^\circ C-207^\circ C)=(36.0\;g)(4.18\; \frac {J}{g^\circ C})(26.0^\circ C-25.0^\circ C)$

Solving for $C_{metal}$, we obtain

$C_{metal}=0.166\; \frac {J} {g^\circ C}$

The specific heat capacity of the unknown metal is 0.166 $\frac {J} {g ^\circ C}$ .

Example 2

To determine the standard enthalpy of the reaction H⁺(aq) + OH^–(aq) → H₂O(l), equal volumes of 0.1 M solutions of HCl and of NaOH can be combined initially at 25°C.

This process is exothermic and as a result, a certain amount of heat q_P will be released into the solution. The number of joules of heat released into each gram of the solution is calculated from the product of the rise in temperature and the specific heat capacity of water (assuming that the solution is dilute enough so that its specific heat capacity is the same as that of pure water's). The total quantity of transferred heat can then be calculated by multiplying the result with the mass of the solution.

$\Delta H=q_P = m_{sol'n}C_{water} \Delta T_{sol'n}$

Note that ΔH = q_P because the process is carried out at constant pressure.