Calorimetry

Calorimetry includes several laboratory techniques for measuring the enthalpy changes or "heats" of chemical reactions. Basically, the measurement of the heat of a reaction involves carrying out the reaction using a known amount of reactant in a calorimeter while measuring the temperature change that occurs in the calorimeter. The calorimeter is insulated to prevent heat exchange with the surroundings and must first be calibrated by carrying out a reaction that has a known enthalpy change.

There are two types of calorimeter: constant pressure and constant volume. An "open" or solution calorimeter operates at the constant pressure of the atmosphere, usually with one or both reactants dissolved in solution. [See p. 185, Brown&LeMay.] The heat measured is qp, which, by definition, gives the enthalpy change, ΔH. A "bomb" calorimeter, on the other hand, operates at constant volume [See p. 186.] and gives qv, from which is obtained the (internal) energy change, ΔE. In practice, the difference between qp and qv is very small, usually less than 0.1 %. Thus the difference between ΔH and ΔE is also small.

There are two basic ideas from thermodynamics that are needed in calorimetry. First, the amount of heat, q, absorbed by a body is proportional to its temperature rise:

q = C × ΔT

where C is the heat capacity of the body. C is a positive quantity that is the ratio of heat absorbed to the temperature rise, C = q / ΔT and is usually given in units of J/K or kJ/K. Note that if the body loses heat (q is negative) then ΔT also will be negative (the temperature falls) and C is a positive quantity.

For example, suppose 50.0 kJ of heat is added to a calorimeter either by electrical heating, chemical reaction, or some other method. If the observed temperature rise is ΔT = +2.50 K, then C = (50.0 kJ) / (2.50 K) = 20.0 kJ/K. If temperatures are recorded in degrees Celsius (°C), then the temperature change, ΔT, is expressed in Kelvins (K) and the heat capacity has units of J/K or kJ/K. However, the unit J/°C or J/deg may still be seen in some older books. If older calorie units are used the conversion is: 1 cal = 4.184 J.

Second, if heat energy is exchanged between two bodies (systems) that are insulated, such that no heat is exchanged with the surroundings, then the heat absorbed by one body (q1) is exactly equal to the heat lost by the second (-q2):

q1 = - q2

[Note: q2 is the heat energy absorbed by the second body; -q2 represents heat lost.]

If q1 is positive - energy gained by the first body - then q2 is negative because the second body lost heat energy. The minus sign is needed in the equation because q1 and q2 have opposite signs. The value of q is always interpreted as heat ABSORBED. That is, heat energy transferred TO the system from the surroundings. A negative value means that heat energy was lost to the surroundings or to another body.

For example, if a hot piece of lead is placed next to a cold piece of steel in an insulated box, heat will transfer from the hot lead to the cold steel until the temperatures are equal. Suppose the amount of thermal energy transferred was 500 J. Then qsteel is +500 J since it absorbed 500 J and qPb is -500 J since it lost the same amount: qsteel = - qPb. The two heats are of equal magnitude but of opposite sign.

Chemical Calorimetry

When a chemical reaction is carried out in a calorimeter, heat energy absorbed by the reacting molecules (qrxn) comes from the calorimeter. Here, the chemical reaction is the system and the calorimeter is the surroundings. Thus a heat transfer takes place between the reaction and the calorimeter:

qrxn = - qcal

Since the heat absorbed by the calorimeter is given by qcal = Ccal × ΔT, the basic equation of chemical calorimetry is obtained:

qrxn = - Ccal × ΔT

or

qp = - Ccal × ΔT              (equation 1)

where qp is the heat absorbed by the reaction at constant pressure. Note that a temperature rise in the calorimeter (ΔT is positive) means that qp is negative. This is the case in an exothermic reaction, where heat energy is given off by the reaction. If the temperature falls (ΔT is negative) then the reaction is endothermic and qp is positive. Equation 1, above, is equivalent to Equations [5.23] and [5.24] on pages 185 and 186 in Brown&LeMay.

In chemical calorimetry, reactions are usually carried out using small quantities of reactant, often in the millimole range. In order to obtain ΔH for mole quantities of reactant, it is necessary to convert qp obtained from the experiment to the molar ΔH value for the reaction. To do this, qp is multiplied by a conversion factor obtained from the amount of reactant actually used and the molar amount indicated in the balanced equation:

             1 mol (of equation)
ΔH = qp × -------------              (equation 2)
             n mol (of reactant)

where n is the number of moles of the limiting reactant used in the calorimeter. ΔH values are usually expressed in units of kJ and are understood to be associated with the numbers of moles indicated in the balanced equation - one mole of "equation."

For example, suppose a calorimeter has a heat capacity of 240 J/K. The following reaction was carried out in the calorimeter using 0.0200 mol of each reactant. The observed temperature change was ΔT = + 4.8 K. (2 significant figures)

NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(liq) DH = ??

First calculate qp, the heat absorbed for the reaction, using equation 1:

qp = - Ccal × T = - (240 J/K) × 4.8 K = - 1152 J

Next, calculate ΔH for one mole using equation 2. The amount of reactant used was 0.0200 mol and the equation requires 1 mole of each reactant. So ΔH is given by:

         1 mol                1 mol
DH = qp × ----- = - 1152 J × ---------- = - 57600 J or - 58 kJ
         n mol              0.0200 mol                  (2 sig. figs.)

In summary, determining ΔH for a reaction requires first that the calorimeter be calibrated to determine the precise value of its heat capacity, Ccal. Then the unknown reaction is carried out and qp is determined from ΔT using equation 1. Finally, ΔH is calculated, using equation 2, from the number of moles of the limiting reactant.

The following example illustrates the two steps of calorimetry: (1) calibration followed by (2) measuring the heat of a reaction.

For example, a calorimeter was calibrated by carrying out the following reaction using 25.0 mL of 2.00 M NaOH and 25.0 mL of 2.00 M HCl. ΔH for the reaction is shown. The temperature of the calorimeter rose from tinit = 25.0 °C to tfinal = 37.5 °C.

NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(liq) DH = -57.3 kJ

Then the following reaction was carried out in the same calorimeter using 105 mg of magnesium metal and the same total volume of solution, 50.0 mL of 1.00 M HCl.

Mg(s) + HCl(aq) --> MgCl2 + H2O(liq) ΔH = ?? kJ

The temperature rose from tinit = 25.0 °C to tfinal = 40.2 °C. What are the heat capacity of the calorimeter (Ccal) and the heat of the reaction (ΔH)?

Solution:

[Calibration] In the first reaction: n = M × V = (2.00 mol/L)×(0.0250 L) = 0.0500 mol. For this reaction: qp = (-57.3 kJ)×(0.0500 mol/1.00 mol) = -2.865 kJ. The temperature change is ΔT = tfinal - tinit = 37.5 °C - 25.0 °C = 12.5 K. So the heat capacity is given by (using equation 1):

C = - qp / ΔT = - (-2.865 kJ) / (12.5 K) = 0.229 kJ/K

[Heat of Reaction] For the magnesium reaction, ΔT = 40.2 °C - 25.0 °C = 15.2 K and qp is:

qp = - (0.229 kJ/K) × (15.2 K) = - 3.48 kJ.

The number of moles used was n = (0.105 g) / (24.3 g/mol) = 0.00432 mol.

Finally, ΔH is obtained as follows:

ΔH = qp / n = (-3.48 kJ) / 0.00432 = - 805 kJ (for one mole of Mg)

Note that magnesium was the limiting reactant here since the amount of HCl was (1.00 M)×(0.100 L) = 0.100 mol, compared to 0.00432 mol of Mg.

Here are a few sample problems based on the above material.

RSS June 13, 2006

© 2006 by Dr. Ronald S. Strange