Hess’s Law

HIGHER GRADE CHEMISTRY CALCULATIONS Hess’s Law

Hess’s Law states that the enthalpy change for a reaction depends only on the enthalpies of the reactants and products and is independent of the route taken for the reaction.

Worked example: Use the enthalpies of combustion in the data booklet to calculate the enthalpy change for: C(s) + 2 H 2 (g)  CH 4 (g) Use the enthalpies of combustion of carbon, hydrogen and methane.

(1) C + O 2 (2) H 2 (3) CH 4 + ½ O + 2O 2 2 -> CO 2   H 2 O CO 2 + 2H 2 O D H 1 D H 2 = -394 kJ.

= -286 kJ.

D H 3 = -882 kJ.

Rearrange the equations to give the one we want.

(1) C + O 2 (2) x 2 2H 2 (3)rev CO 2 + O 2 -> CO  + 2H 2 O  2 2H 2 O CH 4 + 2O 2 D H 1 D H 4 = -394 kJ.

= -572 kJ.

D H 5 = +882 kJ.

Equations cancel out to give the one we want D H = (-394) +(-572) + (+882) = -84 kJ mol -1

Calculations for you to try.

1.

Calculate the enthalpy of formation of ethane using the heats of combustion in the data booklet.

We want D H for 2C + 3H 2  C 2 H 6 Use the enthalpies of combustion of carbon, hydrogen and ethane.

(1) C + O 2 (2) H 2 + ½ O (3) C 2 H 6 2 -> CO 2  H 2 O + 3 ½ O 2  2CO 2 + 3H 2 O D H 1 D H 2 = -394 kJ.

= -286 kJ.

D H 3 = -1560 kJ.

Rearrange the equations to give the one we want.

(1) x 2 2C + 2O 2 (2) x 3 3H 2 + 1 ½ O 2 (3)rev 2CO 2 -> 2 CO  + 3H 2 O  3H 2 O C 2 H 6 2 + 3 ½ O 2 D H 4 = -788 kJ.

D H 5 = -858 kJ.

D H 6 = -1560 kJ.

Adding these equations and cancelling out gives the one we want D H = (-788) +(-858) + (+1560) = -86 kJ mol -1 Higher Grade Chemistry

Calculations for you to try.

2. Calculate the enthalpy of formation of methanol, CH 3 OH, using the heats of combustion in the data booklet.

We want D H for C + 2H 2 + ½ O 2  CH 3 OH Use the enthalpies of combustion of carbon, hydrogen and methanol.

(1) C + O 2 (2) H 2 + ½ O 2 -> CO  (3) CH 3 OH + 1 ½ O 2 2 H 2 O  CO 2 + 2H 2 O D H 1 D H 2 = -394 kJ.

= -286 kJ.

D H 3 = -727 kJ.

Rearrange the equations to give the one we want.

(1) C + O 2 (2) x 2 2H 2 + O 2 (3)rev CO 2 + 2H 2 O -> CO   2 2H 2 O CH 3 OH + 1 ½ O 2 D H 1 D H 4 = -394 kJ.

= -572 kJ.

D H 5 = +727 kJ.

Adding these equations and cancelling out gives the one we want D H = (-394) +(-572) + (+727) = -239 kJ mol -1 Higher Grade Chemistry