Thermodynamics:
Thermochemical Equations

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In this module:
Introduction
Calorimetry
Rules
Hess' Law
DHfº

Calorimetry

Scientists measure the change in thermodynamic quantities in thermochemical equations using a device known as a calorimeter . One kind of calorimeter, known as a coffee cup calorimeter, is shown at left. Coffee cup calorimeters are usually used to measure changes that take place in solution. Since the reaction taking place in the calorimeter is open to the atmosphere, the enthalpy change is measured directly by the device. The insulation provided by the styrofoam ensures that any heat absorbed or released by the system (reactants and products) goes only to the solvent in the cup. In other words, the surroundings are restricted to the solvent in the cup. The thermometer allows the DT of the surroundings to be measured. By the First Law of Thermodynamics (see the Energy module), we know any heat lost by the system must be absorbed by the surroundings, or:

-qsystem = qsurroundings

Using DT, which can be measured, and the heat capacity of the solvent in the coffee cup, the heat lost by the system can be calculated. Since the pressure is constant, this is equal to the enthalpy change for the process.

When 1.00 g of NaOH(s) is dissolved in 100.0 mL of H2O (l) in a coffee cup calorimeter, the temperature of the water rises from 25.00 to 27.66 ºC. The heat capacity of water is 4.184 J/gºC. Find DH for the dissolving process:

NaOH(s) NaOH(aq)      DH = ?

Step 1: Define the system and surroundings. Write your answer in the space below, then click on the check button.


Now compare your answer with the one below.
Please enter your answer in the space at left.

As in most thermochemical equations, the reactants and products, the NaOH(s) and NaOH(aq) in this case, are the best choice for the system. Because of the thermal insulation provided by the styrofoam in the calorimeter, we can confine our definition of the surroundings to the water contained in the calorimeter.

Step 2: Identify and assign signs to all the kinds of energy or work entering or leaving the system. Write your answer in the space below, then click on the check button.

Now compare your answer with the one below.
Please enter your answer in the space at left.

Since the temperature of the water increased, heat must have been transferred from the system to the surroundings. This heat therefore has a negative sign.

Step 3: Predict the units your answer should have. Write your answer in the space below, then click on the check button.

Now compare your answer with the one below.
Please enter your answer in the space at left.

The question is asking for an enthalpy, so the answer should be in joules or kilojoules.

Step 4: Predict the approximate size of your answer. Write your answer in the space below, then click on the check button.

Now compare your answer with the one below.
Please enter your answer in the space at left.

The heat capacity of water is about 4 J/gºC, so the heat transferred should be about 100*4*3 = 1200 J. Since a thermochemical equation relates the heat transferred to moles of reactants or products, this should be expressed as a ratio to the amount of NaOH dissolved, or 1200 J/1 g NaOH * 40 g/mol = 48000 J or 48 kJ.

How much heat is gained by the water in the calorimeter? Click on the correct answer below.
1113 J      11,573 J      11.3 J
Correct! Since -qsystem = qsurroundings, this is also the heat lost by the system.
Remember, DT = Tfinal - Tinitial.
Remember to include the mass of the water in the calorimeter in your calculation.
What is the enthalpy change per mole of NaOH(s)? This will be the DH for the thermochemical equation. Write your answer in the space below and click on the Review Answers button.
Correct!
Remember to round your answer to the correct number of significant figures.
Remember, the system is losing energy, so the sign must be negative.
Remember, the DH must be expressed as a ratio to the number of moles of reactants or products in the thermochemical equation.
That is incorrect, please try again.
The correct answer is -44.5 kJ.

Calorimetry