1036 HW 2 _sol_

1036 HW 2 _sol_ - Chemistry 1036 Homework 2 Solutions 1. a....

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Chemistry 1036 Homework 2 Solutions 1. a. Is the reaction exothermic or endothermic? The reaction is endothermic . You know that because Δ H is a positive value (+90.7 kJ). Energy is added to the system as heat. b. Calculate the amount of heat absorbed when 45.0 g of CH 3 OH are decomposed. Use the fact that for every 1 mole of CH 3 OH reacted, 90.7 kJ of heat are absorbed (from balanced equation above): 33 45.0 g CH OH 1 mol CH OH 90.7 kJ x x = +127 kJ 32.0 g CH OH 1 mol CH OH c. If the enthalpy change is 16.5 kJ, how many grams of hydrogen gas are produced? The balanced equation tells us that the enthalpy change is 90.7 kJ for every 2 moles of H 2 produced: 16.5 kJ x 2 mol H 2 90.7 kJ x 2.02 g H 2 1 mol H 2 = 0.735 g H 2 2. We must manipulate each of the 3 reactions given so that these 3 reactions will add up to give the reaction for which we want Δ H: Reverse Reaction 3 so that there will be 1 mole of C 2 H 4 as a reactant as there is in the reaction we are trying to get. This means that the sign of Δ H for this reaction must be changed from + to -: 3. (Reversed): C 2 H 4 (g) 2C(s) + 2H 2 (g) Δ H = -52.3 kJ (Note negative sign now) Multiply Reaction 2 by 2 so that there will be 2 moles of CF 4 as a product as there are in the reaction we are trying to get. This means that the Δ H value for Reaction 2 must be doubled as well: 2. 2C(s) + 4F 2 (g) 2CF 4 (g) Δ H = 2(-680 kJ) = -1360 kJ Multiply Reaction 1 by 2 so there will be 4 moles of HF as a product as there are in the reaction we are trying to get. This means that the Δ H value for Reaction 1 must be double as well; 1 2H 2 (g) + 2F 2 (g) 4HF(g) Δ H =2( -537 kJ) = -1074 kJ

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Now add Reactions 1, 2, 3 with the above changes: 1 2H 2 (g) + 2F 2 (g) 4HF(g) Δ H =2( -537 kJ) = -1074 kJ 2.
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This homework help was uploaded on 04/10/2008 for the course CHEM 1036 taught by Professor Amateis during the Spring '08 term at Virginia Tech.

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1036 HW 2 _sol_ - Chemistry 1036 Homework 2 Solutions 1. a....

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