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# HESS’S LAW.

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HESS’S LAW

ALSO CALLED … Law of constant heat summation
First law of thermodynamics

HESS’S LAW States that:
The total enthalpy of a reaction is independent of the reaction pathway.

PARAPHRASE OF HESS’S LAW
The heat evolved or absorbed in a chemical process is the same whether the process takes place in one or in several steps. Mathematically, ΔH overall = sum of ΔH’s of individual reactions.

Enthalpy diagram for hess’S law

IMPORTANCE OF HESS’ S LAW
Allows us to calculate the enthalpy changes of reactions that cannot be measured directly in the laboratory.

EXAMPLE In the production of CO2, there are 2 reaction paths that can be taken before final product is obtained. When the enthalpy change of these 2 reactions are added, the enthalpy change of the overall reaction is obtained.

EXAMPLE C + ½ O2  CO H1 = Y kJ CO + ½ O2  CO2 H2 = Z kJ
C + O2  CO2 H3 = XkJ Using Hess’s law, H3 = H1 +H2 H3 = Y kJ + Z kJ H3 = X kJ

MORE QUESTIONS S(s) + 1½O2(g)  SO3(g)
Two reactions occurring in the manufacture of sulfuric acid are shown below: S(s) +O2(g)  SO2(g) ΔH1Ө = –297 kJ SO2(g) + ½O2(g) SO3(g) Δ H2Ө = –92 kJ Deduce the Δ HӨ value of this reaction: S(s) + 1½O2(g)  SO3(g) (November 2005)

RULES FOR SOLVING QUESTIONS ON HESS’S LAW.
When it is necessary to reverse a chemical equation, change the sign of ΔH for that reaction. When multiply equation coefficients, multiply values of ΔH by the same factor. Make sure to rearrange the given equations so that reactants and products are on the appropriate sides of the arrows.

QUESTION 1 Calculate the enthalpy change, ΔH4 for the reaction
C + 2H2 + ½O2  CH3OH ΔH4 Using Hess’s Law, and the following information. CH3OH + 1½O2  CO2 + 2H2O ΔH1 = -676 kJ mol-1 C + O2  CO ΔH2 = -394 kJ mol-1 H2 + ½O2  H2O ΔH3 = -242 kJ mol-1 (May 2006)

QUESTION 2 Using the equations below
Cu(s) + ½ O2(g) → CuO(s)∆Hο = –156 kJmol-1 2Cu(s) + ½ O2(g) → Cu2O(s)∆Hο = –170 kJmol-1 What is the value of ∆Hο (in kJmol-1) for the following reaction? 2CuO(s) → Cu2O(s) + ½ O2(g) A. 142 B. 15 C. –15 D. –142 (May 2004)

QUESTION 3 Consider the following equations.
Mg(s) + ½O2(g) → MgO(s) ∆Hο = –602 kJ H2(g) + ½ O2(g) → H2O(g) ∆Hο = –242 kJ What is the ∆H° value (in kJ) for the following reaction? MgO(s) + H2(g) → Mg(s) + H2O(g) A. –844 B. –360 C. +360 D. +844 (November 2004)

Hess’s law energy cycle

C + ½ O2  CO H = – 110.5 kJ CO + ½ O2  CO2 H = – 283.0 kJ
Reactants H = – kJ CO + ½ O2 Intermediate Enthalpy H = – kJ H = – kJ Note that the states should be included in the diagram. Products CO2

Steps in drawing enthalpy diagrams
Balance the equation(s). Sketch a rough draft based on H values. Draw the overall chemical reaction as an enthalpy diagram with the reactants on one line, and the products on the other line. Draw a reaction representing the intermediate step by placing the relevant reactants on a line. Check arrows. Look at equations to help complete balancing (all levels must have the same number of all atoms). Add axes and H values. NB: Endothermic reactions should have their arrows pointing upward as the enthalpy change is increasing whereas exothermic reactions should have arrows pointing downwards as the enthalpy change is decreasing.

MORE EXAMPLES GeO(s)  Ge(s) + ½ O2(g) H= + 255 kJ
Ge(s) + O2(g)  GeO2(s) H= – kJ GeO(s) + ½ O2(g)  GeO2(s) H= – kJ Ge(s) + O2(g) Intermediate H = kJ GeO(s) + ½ O2(g) – kJ H = Reactants Enthalpy H= – 280 kJ GeO2(s) Products

TO TRY Calculate the enthalpy change for the reaction below
Using Hess’s law and the following information: With the above information, draw an enthalpy diagram for the reaction using the knowledge acquired from the above examples. NO(g) + ½ O2(g)  NO2(g) NO(g)  ½ N2(g) + ½ O2(g) H= – kJ ½ N2(g) + O2(g)  NO2(g) H= kJ

REFERENCE www.chalkbored.com/lessons/chemistry-12/hess-law.ppt

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