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Law of constant heat summation First law of thermodynamics.

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Presentation on theme: "Law of constant heat summation First law of thermodynamics."— Presentation transcript:

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2 Law of constant heat summation First law of thermodynamics

3 States that: The total enthalpy of a reaction is independent of the reaction pathway.

4 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 ΔHs of individual reactions.

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6 Allows us to calculate the enthalpy changes of reactions that cannot be measured directly in the laboratory.

7 In the production of CO 2, 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.

8 C + ½ O 2 CO H 1 = Y kJ CO + ½ O 2 CO 2 H 2 = Z kJ C + O 2 CO 2 H 3 = XkJ Using Hesss law, H 3 = H 1 + H 2 H 3 = Y kJ + Z kJ H 3 = X kJ

9 Two reactions occurring in the manufacture of sulfuric acid are shown below: S(s) +O 2 (g) SO 2 (g)ΔH 1 Ө = –297 kJ SO 2 (g) + ½O 2 (g) SO 3 (g) Δ H 2 Ө = –92 kJ Deduce the Δ H Ө value of this reaction: S(s) + 1½O 2 (g) SO 3 (g) (November 2005)

10 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.

11 Calculate the enthalpy change, ΔH 4 for the reaction C + 2H 2 + ½O 2 CH 3 OH ΔH 4 Using Hesss Law, and the following information. CH 3 OH + 1½O 2 CO 2 + 2H 2 OΔH 1 = -676 kJ mol -1 C + O 2 CO 2 ΔH 2 = -394 kJ mol -1 H 2 + ½O 2 H 2 OΔH 3 = -242 kJ mol -1 (May 2006)

12 Using the equations below Cu(s) + ½ O 2 (g) CuO(s)H ο = –156 kJmol -1 2Cu(s) + ½ O 2 (g) Cu 2 O(s)H ο = –170 kJmol -1 What is the value of H ο (in kJmol -1 ) for the following reaction? 2CuO(s) Cu 2 O(s) + ½ O 2 (g) A.142 B.15 C.–15 D.–142 (May 2004)

13 Consider the following equations. Mg(s) + ½O 2 (g) MgO(s) H ο = –602 kJ H 2 (g) + ½ O 2 (g) H 2 O(g) H ο = –242 kJ What is the H° value (in kJ) for the following reaction? MgO(s) + H 2 (g) Mg(s) + H 2 O(g) A.–844 B.–360 C.+360 D.+844 (November 2004)

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15 C + O 2 CO 2 H = – kJ Reactants Intermediate Products C + O 2 CO 2 CO Enthalpy H = – kJ H = – kJ H = – kJ + ½ O 2 C + ½ O 2 CO H = – kJ CO + ½ O 2 CO 2 H = – kJ

16 1. Balance the equation(s). 2. Sketch a rough draft based on H values. 3. Draw the overall chemical reaction as an enthalpy diagram with the reactants on one line, and the products on the other line. 4. Draw a reaction representing the intermediate step by placing the relevant reactants on a line. 5. Check arrows. 6. Look at equations to help complete balancing (all levels must have the same number of all atoms). 7. Add axes and H values.

17 GeO(s) Ge(s) + ½ O 2 (g) H = kJ Ge(s) + O 2 (g) GeO 2 (s) H = – kJ GeO(s) + ½ O 2 (g) GeO 2 (s) Reactants Products Intermediate GeO(s) + ½ O 2 (g) GeO 2 (s) Ge(s) + O 2 (g) Enthalpy H = kJ H = – kJ H = – 280 kJ H = – kJ

18 Calculate the enthalpy change for the reaction below Using Hesss 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 O 2 (g) NO 2 (g) NO(g) ½ N 2 (g) + ½ O 2 (g) H = – kJ ½ N 2 (g) + O 2 (g) NO 2 (g) H = kJ

19 law ml ml hapter9section5.rhtml hapter9section5.rhtml istry/calculations_3/hess_rev1.shtml istry/calculations_3/hess_rev1.shtml ng_IB/player.html ng_IB/player.html


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