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Energy Lecture 4 Hess’s Law & Review

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**Calculating Enthalpy Change**

A theoretical way to determine ∆H for a chemical reaction is provided by Hess’s law, which states that if two or more thermochemical equations can be added to produce a final equation for a reaction, then the enthalpy change for the final reaction equals the sum of the enthalpy changes for the individual reactions.

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Hess’s Law There is an amount of heat associated with every chemical reaction.

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Hess’s Law Often you know the heat for parts of the reaction, and you must add them together to find the heat for the total reaction.

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Applying Hess’s Law Use 2 thermochemical reactions to determine ∆H for the oxidation of ethanol (C2H5OH) to form acetaldehyde (C2H4O) and water. Here is the overall reaction: ethanol + oxygen gas acetaldehyde + water Here are the two component reactions:

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Applying Hess’s Law For the overall reaction, Acetaldehyde should be on the right side of the equation, so reverse equation a. Note that you must change the sign of ∆H. The desired equation has two moles of ethanol, so double equation b and its ∆H.

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Applying Hess’s Law Add these two equations, and cancel any terms common to both sides of the combined equation. 1 2 ∆ H = -349kJ Note that ∆H is negative, which means the reaction is exothermic. (releasing energy)

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**Basic Assessment Questions**

Practice Hess’s Law! Use reactions a and b to determine ∆H for this single-displacement reaction. Cl2(g) + 2HBr(g) HCl (g) + Br2(g) a. H2(g) + Cl2(g) HCl (g) ∆H= -185 b. H2(g) + Br2(g) HBr (g) ∆H= -73

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**Practice Hess’s Law a. H2(g) + Cl2(g) 2HCl (g) ∆H= -185kJ**

Keep equation “a” as written because HCl is on the right in the total reaction: a. H2(g) + Cl2(g) HCl (g) ∆H= -185kJ Flip equation “b” because HBr needs to be on the left in the overall equation b. H2(g) + Br2(g) HBr (g) ∆H= -73kJ b HBr (g) H2(g) + Br2(g) ∆H=73kJ

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**a. H2(g) + Cl2(g) 2HCl (g) ∆H= -185 kJ**

Now add the two equations together a. H2(g) + Cl2(g) HCl (g) ∆H= -185 kJ b HBr (g) H2(g) + Br2(g) ∆H= 73kJ Cl2(g) + 2HBr(g) HCl (g) + Br2(g) ∆H = -112 = -112kJ

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Answer ∆Hrxn = –112 kJ

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Practice using q=cm∆T A 15.6-g sample of ethanol absorbs 868 J as it is heated. If the initial temperature of the ethanol was 21.5°C, what is the final temperature of the ethanol? Hint: solve for ∆T then add 21.5!

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**A 15. 6-g sample of ethanol absorbs 868 J as it is heated**

A 15.6-g sample of ethanol absorbs 868 J as it is heated. If the initial temperature of the ethanol was 21.5°C, what is the final temperature of the ethanol? 868 J (2.44J/gºC ) (15.6g) ∆T = q . c m = Remember this is not your final answer. You are looking for the final temp… so add the initial temp to this number ∆T= 22.8ºC 22.8ºC ºC = 44.3ºC = final temp

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Practice q=mc∆T again! If 335 g water at 65.5°C loses 9750 J of heat, what is the final temperature of the water?

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Answer 58.5°C

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**Practice using q=mol x ∆H**

How much heat is evolved when 24.9 g of propanol (C3H7OH) is burned? ∆Hcomb = –2010 kJ/mol Molar mass of C3H7OH is 60.1g/mol

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Answer 833 kJ

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Molar Enthalpy Recall that when we write a thermochemical equation the coefficients represent moles of particles Therefore, 1 H 2(g) + ½ O 2(g) 1 H 2.

Molar Enthalpy Recall that when we write a thermochemical equation the coefficients represent moles of particles Therefore, 1 H 2(g) + ½ O 2(g) 1 H 2.

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