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5.3 Enthalpy  (Page 8/25)

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Before we further practice using Hess’s law, let us recall two important features of Δ H .

  1. Δ H is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of NO 2 ( g ) is +33.2 kJ:

    1 2 N 2 ( g ) + O 2 ( g ) ⟶ NO 2 ( g ) Δ H = +33.2 kJ

    When 2 moles of NO 2 (twice as much) are formed, the Δ H will be twice as large:

    N 2 ( g ) + 2 O 2 ( g ) ⟶ 2 NO 2 ( g ) Δ H = +66.4 kJ

    In general, if we multiply or divide an equation by a number, then the enthalpy change should also be multiplied or divided by the same number.

  2. Δ H for a reaction in one direction is equal in magnitude and opposite in sign to Δ H for the reaction in the reverse direction. For example, given that:

    H 2 ( g ) + Cl 2 ( g ) ⟶ 2 HCl ( g ) Δ H = −184.6 kJ

    Then, for the “reverse” reaction, the enthalpy change is also “reversed”:

    2 HCl ( g ) ⟶ H 2 ( g ) + Cl 2 ( g ) Δ H = +184.6 kJ

Stepwise calculation of Δ H f ° Using hess’s law

Determine the enthalpy of formation, Δ H f ° , of FeCl 3 ( s ) from the enthalpy changes of the following two-step process that occurs under standard state conditions:

Fe ( s ) + Cl 2 ( g ) ⟶ FeCl 2 ( s ) Δ H ° = −341.8 kJ
FeCl 2 ( s ) + 1 2 Cl 2 ( g ) ⟶ FeCl 3 ( s ) Δ H ° = −57.7 kJ

Solution

We are trying to find the standard enthalpy of formation of FeCl 3 ( s ), which is equal to Δ H ° for the reaction:

Fe ( s ) + 3 2 Cl 2 ( g ) ⟶ FeCl 3 ( s ) Δ H f ° = ?

Looking at the reactions, we see that the reaction for which we want to find Δ H ° is the sum of the two reactions with known Δ H values, so we must sum their Δ H s:

Fe ( s ) + Cl 2 ( g ) ⟶ FeCl 2 ( s ) Δ H ° = −341.8 kJ FeCl 2 ( s ) + 1 2 Cl 2 ( g ) ⟶ FeCl 3 ( s ) Fe ( s ) + 1 2 Cl 2 ( g ) ⟶ FeCl 3 ( s ) Δ H ° = −57.7 kJ Δ H ° = −399.5 kJ

The enthalpy of formation, Δ H f ° , of FeCl 3 ( s ) is −399.5 kJ/mol.

Check your learning

Calculate Δ H for the process:

N 2 ( g ) + 2 O 2 ( g ) ⟶ 2 NO 2 ( g )

from the following information:

N 2 ( g ) + O 2 ( g ) ⟶ 2 NO ( g ) Δ H = 180.5 kJ
NO ( g ) + 1 2 O 2 ( g ) ⟶ NO 2 ( g ) Δ H = −57.06 kJ

Answer:

66.4 kJ

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Here is a less straightforward example that illustrates the thought process involved in solving many Hess’s law problems. It shows how we can find many standard enthalpies of formation (and other values of Δ H ) if they are difficult to determine experimentally.

A more challenging problem using hess’s law

Chlorine monofluoride can react with fluorine to form chlorine trifluoride:

(i) ClF ( g ) + F 2 ( g ) ⟶ ClF 3 ( g ) Δ H ° = ?

Use the reactions here to determine the Δ H ° for reaction (i) :

(ii) 2 OF 2 ( g ) ⟶ O 2 ( g ) + 2 F 2 ( g ) Δ H ( i i ) ° = −49.4 kJ

(iii) 2 ClF ( g ) + O 2 ( g ) ⟶ Cl 2 O ( g ) + OF 2 ( g ) Δ H ( i i i ) ° = +205.6 kJ

(iv) ClF 3 ( g ) + O 2 ( g ) ⟶ 1 2 Cl 2 O ( g ) + 3 2 OF 2 ( g ) Δ H ( i v ) ° = +266.7 kJ

Solution

Our goal is to manipulate and combine reactions (ii) , (iii) , and (iv) such that they add up to reaction (i) . Going from left to right in (i) , we first see that ClF( g ) is needed as a reactant. This can be obtained by multiplying reaction (iii) by 1 2 , which means that the Δ H ° change is also multiplied by 1 2 :

ClF ( g ) + 1 2 O 2 ( g ) ⟶ 1 2 Cl 2 O ( g ) + 1 2 OF 2 ( g ) Δ H ° = 1 2 ( 205.6 ) = +102.8 kJ

Next, we see that F 2 is also needed as a reactant. To get this, reverse and halve reaction (ii) , which means that the Δ H ° changes sign and is halved:

1 2 O 2 ( g ) + F 2 ( g ) ⟶ OF 2 ( g ) Δ H ° = +24.7 kJ

To get ClF 3 as a product, reverse (iv) , changing the sign of Δ H °:

1 2 Cl 2 O ( g ) + 3 2 OF 2 ( g ) ⟶ ClF 3 ( g ) + O 2 ( g ) Δ H ° = −266.7 kJ

Now check to make sure that these reactions add up to the reaction we want:

ClF ( g ) + 1 2 O 2 ( g ) ⟶ 1 2 Cl 2 O ( g ) + 1 2 OF 2 ( g ) Δ H ° = +102.8 kJ 1 2 O 2 ( g ) + F 2 ( g ) ⟶ OF 2 ( g ) Δ H ° = +24.7 kJ 1 2 Cl 2 O ( g ) + 3 2 OF 2 ( g ) ⟶ ClF 3 ( g ) + O 2 ( g ) ClF ( g ) + F 2 ⟶ ClF 3 ( g ) Δ H ° = −266.7 kJ Δ H ° = −139.2 kJ

Reactants 1 2 O 2 and 1 2 O 2 cancel out product O 2 ; product 1 2 Cl 2 O cancels reactant 1 2 Cl 2 O; and reactant 3 2 OF 2 is cancelled by products 1 2 OF 2 and OF 2 . This leaves only reactants ClF( g ) and F 2 ( g ) and product ClF 3 ( g ), which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified Δ H ° values will give the desired Δ H °:

Δ H ° = ( +102.8 kJ ) + ( 24.7 kJ ) + ( −266.7 kJ ) = −139.2 kJ

Check your learning

Aluminum chloride can be formed from its elements:

(i) 2 Al ( s ) + 3 Cl 2 ( g ) ⟶ 2 AlCl 3 ( s ) Δ H ° = ?

Use the reactions here to determine the Δ H ° for reaction (i) :

(ii) HCl ( g ) ⟶ HCl ( a q ) Δ H ( i i ) ° = −74.8 kJ

(iii) H 2 ( g ) + Cl 2 ( g ) ⟶ 2 HCl ( g ) Δ H ( i i i ) ° = −185 kJ

(iv) AlCl 3 ( a q ) ⟶ AlCl 3 ( s ) Δ H ( i v ) ° = +323 kJ/mol

(v) 2Al ( s ) + 6 HCl ( a q ) ⟶ 2 AlCl 3 ( a q ) + 3 H 2 ( g ) Δ H ( v ) ° = −1049 kJ

Answer:

−1407 kJ

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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