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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.

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Presentation on theme: "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."— Presentation transcript:

1 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 -> CO 2  H 1 = -394 kJ. (2)H 2 + ½ O 2  H 2 O  H 2 = -286 kJ. (3) CH 4 + 2O 2  CO 2 + 2H 2 O  H 3 = -882 kJ. Rearrange the equations to give the one we want. (1) C + O 2 -> CO 2  H 1 = -394 kJ. (2) x 2 2H 2 + O 2  2H 2 O  H 4 = -572 kJ. (3)rev CO 2 + 2H 2 O  CH 4 + 2O 2  H 5 = +882 kJ. Equations cancel out to give the one we want  H = (-394) +(-572) + (+882) = -84 kJ mol -1

2 Higher Grade Chemistry Calculations for you to try. 1.Calculate the enthalpy of formation of ethane using the heats of combustion in the data booklet. We want  H for 2C + 3H 2  C 2 H 6 Use the enthalpies of combustion of carbon, hydrogen and ethane. (1)C + O 2 -> CO 2  H 1 = -394 kJ. (2)H 2 + ½ O 2  H 2 O  H 2 = -286 kJ. (3) C 2 H ½ O 2  2CO 2 + 3H 2 O  H 3 = kJ. Rearrange the equations to give the one we want. (1) x 2 2C + 2O 2 -> 2 CO 2  H 4 = -788 kJ. (2) x 3 3H ½ O 2  3H 2 O  H 5 = -858 kJ. (3)rev 2CO 2 + 3H 2 O  C 2 H ½ O 2  H 6 = kJ. Adding these equations and cancelling out gives the one we want  H = (-788) +(-858) + (+1560) = -86 kJ mol -1

3 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  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 -> CO 2  H 1 = -394 kJ. (2)H 2 + ½ O 2  H 2 O  H 2 = -286 kJ. (3) CH 3 OH + 1 ½ O 2  CO 2 + 2H 2 O  H 3 = -727 kJ. Rearrange the equations to give the one we want. (1) C + O 2 -> CO 2  H 1 = -394 kJ. (2) x 2 2H 2 + O 2  2H 2 O  H 4 = -572 kJ. (3)rev CO 2 + 2H 2 O  CH 3 OH + 1 ½ O 2  H 5 = +727 kJ. Adding these equations and cancelling out gives the one we want  H = (-394) +(-572) + (+727) = -239 kJ mol -1


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