**How to Measure ΔH _{rxn}: 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 ΔH_{rxn }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 ΔH_{rxn }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 C_{s} (specific heat) is typically expressed in terms of J /^{o}C× g. TheC_{s} some common substances include: H_{2}O - 4.18 J/ ^{o}C× g, Au- .128 J/ ^{o}C× g, Fe - J/ ^{o}C × g. Therefore, when calculating heat evolution of a particular substance we use the next equation.

- q = m (mass of substance) × C
_{s }× ΔT

Example: A copper penny (C_{s }of Cu = .385 J/^{o}C×g) is left in the snow, which is -8.0^{o}C. If the penny weighs 3.10 g, how much heat will the penny absorb as it warms a human body temperature of 37.0 ^{o}C? Given: m = 3.10 g C_{s }= .385 J/ ^{o}C × g ΔT = (T_{f} – T_{i}) = 45.0 ^{o}C

q= m × C_{s }× ΔT -----> q = 3.10g × .385 J/ ^{o}C × g × 45^{o}C -----> 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 ΔH_{rxn. }Remember that in order to determine the actual amount of heat evolved, we must determine the number of calories absorbed by coffee cup (C_{cal}) in order to calculate the actual ΔH_{rxn}. This involves the use of three equations:

- q
_{sol}= m_{sol }(in grams) × C_{s},_{sol}× Δ T - q
_{sol =}- q_{rxn} - ΔH
_{rxn.}= q_{rxn}/ 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 25.6^{o}C to 32.8^{o}C. Given C_{s, sol }= 4.18 J/^{ o}C × g, d_{sol} (density of solution) = 1.00 g/ml. 100.0 mL sol. × 1 g/ 1 mL = 1.00 × 10 ^{3 }g

Mg _{(s)} + 2HCl _{(aq)} -----------> MgCl_{2 (aq)} + H_{2 (g)}

q_{sol} = m_{sol} × C_{s}, _{sol} × Δ T -----> q_{sol }= 1.00 × 10 ^{3 }g × 4.18 J/^{ o}C × g × 7.2^{ o}C -----> 3.0× 10 ^{3 }J

q_{sol }= - q_{rxn} = -3.0 × 10 ^{3 }J -----> ΔH_{rxn.}= q_{rxn}/ m (in moles) -----> .158g Mg × 1 mol Mg/ 24.31 g

Mg = 6.49 × 10 ^{-3} mol Mg -----> ΔH_{rxn.}= q_{rxn}/ m (in moles) -----> ΔH_{rxn }= -3.0 × 10 ^{3 }J/ 6.49 ×

10 ^{-3} mol

Therefore the ΔH_{rxn =} - 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 ΔH_{rxn =} - 4.6 × 10 ^{5 }J.

Advanced Problem: A scientist wishes to calculate the C_{cal} 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 C_{cal} for the calorimeter?

A. q_{hot} = m × C_{cal} × Δ T = 50.0 × 4.184 × -28 = -5860 J

B. q_{cold} = m × C_{cal} × ΔT = 50.0 × 4.184 × 19 = 3975 J

C. q_{hot }+ q_{cold } + q_{cal }= 0 -----> q_{cal} = 1885 J

D. C_{cal } = q_{cal}/ΔT (from Equation 1) = 1885J/ 19K -----> C_{cal} = 99 J/K

© 2012 Emilee Leach