How to Measure ΔHrxn: Constant-Pressure Calorimetry

Thermochemistry is a branch of the field that many students find exceptionally challenging when taking their first chemistry class, particularly in regard the measure of the ΔHrxn in constant-pressure calorimetry. Because these measurements require that one complete several equations in order to find the solution, this will provide a step-by-step guide to setting up and solving these types of equations. To begin, let’s first review some important equations that may be required to measure the ΔHrxn of a reaction:

• q (heat) = C (heat capacity) × ΔT (temperature change)

The heat capacity of a system must be determined as substances have a higher heat capacity will therefore illustrate a smaller change in temperature than those with a low heat capacity. The Cs (specific heat) is typically expressed in terms of J /oC× g. TheCs some common substances include: H2O - 4.18 J/ oC× g, Au- .128 J/ oC× g, Fe - J/ oC × g. Therefore, when calculating heat evolution of a particular substance we use the next equation.

• q = m (mass of substance) × Cs × ΔT

Example: A copper penny (Cs of Cu = .385 J/oC×g) is left in the snow, which is -8.0oC. If the penny weighs 3.10 g, how much heat will the penny absorb as it warms a human body temperature of 37.0 oC? Given: m = 3.10 g Cs = .385 J/ oC × g ΔT = (Tf – Ti) = 45.0 oC

q= m × Cs × ΔT   ----->  q = 3.10g × .385 J/ oC × g × 45oC   ----->  q = 53.7 J

Therefore, the answer is that the penny would absorb 53.7 J of heat.

How to use this equation when using a coffee calorimeter to measure ΔHrxn. Remember that in order to determine the actual amount of heat evolved, we must determine the number of calories absorbed by coffee cup (Ccal) in order to calculate the actual ΔHrxn. This involves the use of three equations:

• qsol = msol (in grams) × Cs, sol × Δ T
• qsol = - qrxn
• ΔHrxn.= qrxn/ m (in moles)

Example: .158 g of Mg is combined with HCL to make 100.0 mL of solution in a coffee-cup calorimeter. The temperature rises from 25.6oC to 32.8oC. Given Cs, sol = 4.18 J/ oC × g, dsol (density of solution) = 1.00 g/ml. 100.0 mL sol. × 1 g/ 1 mL = 1.00 × 10 3 g

Mg (s) + 2HCl (aq) -----------> MgCl2 (aq) + H2 (g)

qsol = msol × Cs, sol × Δ T -----> qsol  =  1.00 × 10 3 g  × 4.18 J/ oC × g × 7.2 oC  -----> 3.0× 10 3 J

qsol = - qrxn = -3.0 × 10 3 J -----> ΔHrxn.= qrxn/ m (in moles)  -----> .158g Mg × 1 mol Mg/ 24.31 g

Mg = 6.49 × 10 -3 mol Mg ----->  ΔHrxn.= qrxn/ m (in moles)  ----->  ΔHrxn =  -3.0 × 10 3 J/ 6.49 ×

10 -3 mol

Therefore the ΔHrxn = - 4.6 × 10 5 J/mol Mg. Looking at the balanced equation for this reaction since there only one mole of Mg in the equation (as opposed to the 2 mol HCl), we can determine that the ΔHrxn = - 4.6 × 10 5 J.

Advanced Problem: A scientist wishes to calculate the Ccal of his coffee calorimeter. He pours 50.0 mL of water at 345 K into the calorimeter which contains 50.0 mL of water at 298 K. The final temperature is recorded to be 317K. What is the Ccal for the calorimeter?

A. qhot = m × Ccal × Δ T = 50.0 × 4.184 × -28 = -5860 J

B. qcold =  m × Ccal × ΔT = 50.0 × 4.184 × 19 = 3975 J

C. qhot  +  qcold  + qcal = 0   ----->   qcal = 1885 J

D. Ccal  = qcal/ΔT (from Equation 1) = 1885J/ 19K ----->  Ccal = 99 J/K