Humboldt State University ® Department of Chemistry

Richard A. Paselk

	General Chemistry	Spring 2011
Exercise: Thermal Chemistry	© R. Paselk 2008
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Thermal Chemistry

Hess's Law & Enthalpy of Formation Problem Answers

1. Find the value of H for the reaction below at 25°C:

NH_3(g) + HCl_(g) NH₄Cl_(s)

From Table of heats of formation find H°_(f) values:

Tables often just list compounds (e.g. NH₄Cl_(s)) and H°_(f) since known to be from elements.

1/2 N2(g) + 3/2 H2(g) NH3(g)	H°(f) = -46.1 kJ mol-1
1/2 H2(g) + 1/2 Cl2(g) HCl(g)	H°(f) = -92.3 kJ mol-1
1/2 N2(g) + 1/2 H2 + 1/2 Cl2(g) NH4Cl(s)	H°(f) = -314.4 kJ mol-1

Looking at the original equation, NH_3(g) + HCl_(g) NH₄Cl_(s) the first two formation reactions must be reversed, which changes the sign of the heats of formation:

NH3(g) 1/2 N2(g) + 3/2 H 2(g)	H°(f) = +46.1 kJ mol-1
HCl(g)  1/2 H2(g) + 1/2 Cl2(g)	H°(f) = +92.3 kJ mol-1
1/2 N2(g) + 1/2 H2 + 1/2 Cl2(g) NH4Cl(s)	H°(f) = -314.4 kJ mol-1
cancelling items which appear on both sides and adding the equations we then get:
NH3(g) + HCl(g) NH4Cl(s)	H°(f) = -176.0 kJ mol-1

2. Calculate the value of H for the reaction below at 25°C:

Na₂O_(s) + H₂O_(l) 2 NaOH_(s)

From Table of heats of formation find H°_(f) values:

Na(s) + 1/2 O2(g) + 1/2 H2(g) NaOH(s)	H°(f) = -426.7 kJ mol-1
H2(g) + 1/2 O2(g) H2O(l)	H°(f) = -285.8 kJ mol-1 (note that the state of water must be specified as tables may have gas, liquid & solid)
2 Na(s) + O2(g) Na2O(s)	H°(f) = -415.9 kJ mol-1

Looking at the original equation, Na₂O_(s) + H₂O_(l) 2 NaOH_(s) the first two formation reactions must be reversed, which changes the sign of the heats of formation.

Instead of wriing thte reactions in reverse we can simple put the reactions together, adding the H°_(f)for products, and subtracting H°_(f) for reactants (since the reaction directions are reversed), and multiplying H°_(f) values by the coefficients of the balanced equation (since all of the formation reactions were based on coefficients of one).

H°_rxn = (H°_(f))_prod - (H°_(f))_react

H_rxn = 2 (H_NaOH) - (H_Na2O + H_H2O)

First we need to write a balanced equation for the reaction:

CH₃OH_(l) + 3/2 O_2(g) CO_2(g) + 2 H₂O_(g)

The equation for the heat is then:

H°_rxn = (H°_(f))_prod - (H°_(f))_react

where the summation () includes the coeficients for each reactant and product

From the table of heats of formation we find:

Substance	H°(f)kJ/mol
CH3OH(l)	-239.0
CO2(g)	-393.5
H2O(g)	-241.8

There is no value for oxygen since it is in its elemental form, which is by definition zero.

Plugging these values into our equation we get:

H°_rxn = [(H°_(f))_CO2 + 2(H°_(f))_H2O ] - [(H°_(f))_CH3OH+ 3/2(H°_(f))_O2]

H°_rxn = [-393.5 + 2(-241.8)] - [-239.0 + 3/2(0)]

H°_rxn = -638.1 kJ

 

In addition to these exercises you should familiarize yourself with the text materials referenced below.

• Sections 6.3-6.4 (pp 242-252), in your textbook-you should be able to do the examples and exercises in the assigned problems listed on the Schedule.

Thermal Chemistry Module

© R A Paselk

Last modified 2 March 2008