Question

You are interested in determining the heat of isomerization of two isomers (A and B), so you measure their heats of combustion. The difference between the heats of combustion is equal to the heat of isomerization. You are given two separate samples of (1.5 ± 0.1) mol for each isomer, A and B, and they are burned in a constant-pressure calorimeter.

The heat of capacity of the calorimeter is (8.44 ± 0.05) kJ K-1. Combustion of isomer A causes the temperature to rise by (2.0 ± 0.1) K, and isomer B causes the temperature to rise by (3.5 ± 0.1) K.

(a) Calculate the molar heat of isomerization (ΔHi = ΔHB - ΔHA)

(b) Calculate the uncertainty in the molar heat of isomerization. Assume that the errors in the moles of isomers A and B, heat capacity of the calorimeter, and temperature readings are small and independent.

Answer 1

a) molar heat of isomerization =ΔHi = ΔHB - ΔHA

with actual measurements(ignoring uncertainty in values)

ΔHA=8.44 Kj k-1 * 2.0 k/1.5 mol=11.25 KJ /mol

ΔHB=8.44 Kj k-1 * 3.5 k/1.5 mol=19.69 KJ/mol

ΔHi=19.69-11.25=8.44 KJ/mol

b) considering the uncertainty in measured quantities (actual valueuncertainty) we can calculate the maximum and minimum values.

minimum heat of capacity of the calorimeter is (8.44 - 0.05) kJ K-1=8.39kJ K-1

minimum temperature rise for A (2.0 -0.1) K=1.9 K

minimum moles of A=(1.5 - 0.1) mol=1.4 mol

ΔHA=8.39 Kj k-1 * 1.9 k/1.4 mol= 11.39 KJ /mol

taking maximum values,

heat of capacity of the calorimeter is (8.44 + 0.05) kJ K-1=8.49kJ K-1

temperature rise for A (2.0 +0.1) K=2.1 K

moles of A=(1.5 +0.1) mol=1.6 mol

ΔHA=8.49 Kj k-1 * 2.1 k/1.6 mol=11.14 KJ /mol

So ΔHA=11.25 0.12 kj/mol

similarly we can find

ΔHB=19.69 KJ/mol(actual)

ΔHB (with minimum values)=8.39 Kj k-1 * 3.4 k/1.4 mol=20.38 KJ /mol

ΔHB (with max values)=8.49 Kj k-1 * 3.6 k/1.6 mol=19.10 KJ /mol

uncertainty=20.38 KJ /mol-19.10 KJ /mol/2=0.64

so ΔHB=19.690.64 KJ/mol

finally ,ΔHi=(19.690.64 )-(11.25 0.12)

=8.44 uncertainty in values of ΔHB - ΔHA=8.440.52 KJ/mol