Chemistry, Third edition

13 · ENERGY CHANGES IN CHEMICAL REACTIONS

Calculation of H

—

using standard enthalpies of

formation

On page 222, we used Hess’s law to calculate H^ —for the reaction

HCl(g)NH 3 (g)NH 4 Cl(s)

obtainingH—^  176.0 kJ mol^1. The same answer may be obtained by adding

the standard enthalpies of formation of HCl(g) and NH 3 (g) together, and subtract-

ing this sum from the standard enthalpy of formation of ammonium chloride:

H—^ H—f^ (NH 4 Cl(s))[Hf—^ (HCl(g))H^ —f(NH 3 (g))]

TheH^ —f values (kJ mol^1 ) obtained from Table 13.2 are:

H—f^ (NH 4 Cl(s)) 314.4; Hf—^ (HCl(g)) 92.3; Hf—^ (NH 3 (g))] 46.1

so that

H—^ (314.4)(92.3() 46.1)  176.0 kJ mol^1

This method is easier to use than the algebraic method based upon Hess’s law, and

we will now generalize the calculation so that it applies to all reactions:

●To work out the standard enthalpy change (H^ —) of any reaction from H—f^ f

values, substitute into the following equation:

H—^ H—f^ (products)H—f^ (reactants) (13.15)

whereH—f^ (products) is the sum of the standard enthalpies of formation of the

products in the reaction, and H—f^ (reactants) is the sum of the standard

enthalpies of formation of the reactants.

●When summing the enthalpies of formation, the enthalpy of formation of each

substance must be multiplied by its coefficientin the balanced equation for the

reaction. This will be seen in Example 13.4.

228

Example 13.4

CalculateH^ —at 298 K for the production of three moles of

copper by the reaction:

3CuO(s)2NH 3 (g) N 2 (g)3H 2 O(l)3Cu(s) (A)

Answer

H^ —H^ —f(products)H^ —f(reactants)

H^ —[H—f^ (N 2 (g)) 3 H—f^ (H 2 O(l)) 3 Hf^ —(Cu(s))][3H—f^ (CuO(s)) 2 H^ —f(NH 3 (g))]

  

The numbers indicated by arrows are the coefficients in the chemical equation.

Using the standard enthalpy of formation data in Table 13.2,

Enthalpy of formation

Complete the following thermochemical equations which refer to the formation of one mole

of the substances shown at 298 K:

................ . H 2 SO 4 (l) H—^ f 814kJmol^1
................ . C 6 H 12 O 6 (s) H—^ f 1268 kJ mol^1

Exercise 13G