Lecture Notes C: Thermodynamics I (cont)

Transcription

1 Lecture Notes C: Thermodynamics I (cont) 1) The tools of thermochemistry (from page 2 of Lecture Notes B) How do we measure and quantify heat given off or absorbed by a reaction (lecture notes B)? Definition of enthalpy How do we predict the amount of heat that will be given off? What is the correct way to summarize the results of experiments so we can predict new situations? Standard states and heats of formation (these notes) How can we predict it for reactions we haven t seen before? Bond enthalpies (next notes) 2) Standard states Consider listing the different in altitude between all cities: Pittsburgh Scranton NYC Denver Pittsburgh Scranton NYC Denver As opposed to a table of altitude above sea level (in feet) Pittsburgh 743 NYC 54 Denver 5183 Scranton 754 Sea level acts as a standard state sea level ocean altitude B altitude A altitude difference between A and B is difference between altitudes relative to sea level Altitude(Pittsburgh Scranton) = Altitude(Scranton)-Altitude(Pittsburgh) 11 = We can do a similar thing with reaction enthalpies. Instead of listing the reaction enthalpy of all reactions: C 2 H 2 (g) + 5/2 O 2 (g) H 2 O (g) + 2 CO 2 (g) H = kj/mol Heats of formation serve the role of altitude above sea level H f o (C 2 H 2(g) ) = kj/mol H f o (CO 2(g) ) = kj/mol H f o (H 2 O (l) ) = kj/mol H f o (O 2(g) ) = 0 kj/mol H(reaction) = 2 H f o (CO 2(g) ) + H f o (H 2 O (l) ) - H f o (C 2 H 2(g) ) 5/2 H f o (O 2(g) ) = 2 ( ) + ( ) (226.7) 5/2 (0) = kj Lecture Notes C: Thermodynamics I(cont) Distributed on: Wednesday, January 19, 2005 Page 1 of 7

2 So what is the analogue of sea level for chemicals? Chemical Standard States gas: 1 atm and 25 o C substance in aqueous solution: 1M concentration element or compound: most stable form at 1 atm and 25 o C (O 2, H 2, graphite (C) etc.) 3) Heats of formation (or Enthalpy s of formation) Enthalpy change associated with creating one mole of a chemical substance from the elements in their standard states, H 2 (g) + ½ O 2 (g) H 2 O (l) H o = kj H f o ( H 2 O(l) ) = kj/mole The heat of formation of FeCO 3 (s) is kj/mole. Write the corresponding reaction and give its reaction enthalpy. 4) Calculating reaction enthalpies from heats of formation. Elements in standard states c H f o (C) + d H f o (D) -a H f o (A) - b H f o (B) products reactants cc dd aa bb aa + bb cc + dd Lecture Notes C: Thermodynamics I(cont) Distributed on: Wednesday, January 19, 2005 Page 2 of 7

3 Using the data in the appendix of the textbook, calculate the enthalpy of combustion for methanol vapor (CH 3 OH (g) ). Assume the combustion produces CO 2(g) and H 2 O (g). H f o (CH 3 OH (g) ) = kj/mol ; H f o (CO 2(g) ) = kj/mol ; H f o (H 2 O (g) ) = kj/mol C (graphite) 2 O 2 (g) 2 H 2 (g) CH 3 OH(g) 3/2 O 2 (g) CO 2 (g) 2 H 2 O (g) This is a result of Hess s law H o f ( CO 2(g) ) C (graphite) + O 2(g) CO 2 (g) +2 H o f ( H 2 O (g) ) 2 H 2(g) + O 2 (g) 2 H 2 O (g) H o f ( CH 3 OH (g) ) CH 3 OH (g) C (graphite) + 2 H 2 (g) + ½ O 2 (g) 3/2 H o f ( O 2(g) ) 3/2 O 2 (g) 3/2 O 2 (g) H CH 3 OH (g) + 3/2 O 2 (g) 2 H 2 O (g) + CO 2 (g) Lecture Notes C: Thermodynamics I(cont) Distributed on: Wednesday, January 19, 2005 Page 3 of 7

4 Concept Which of the following is correct? a) H f o (CS 2 (l)) = kj/mol ; H f o (CS 2 (g)) = kj/mol b) H f o (CS 2 (l)) = kj/mol ; H f o (CS 2 (g)) = kj/mol Concept State whether each of the following is obviously incorrect: 1) H f o (N 2 (g) at 25 o C ) = kj/mole a) obviously incorrect b) could be ok 2) H f o (C 2 H 4 (g) at 25 o C) = kj/mole a) obviously incorrect b) could be ok 3) H f o (Si (g) at 25 o C ) = kj/mole a) obviously incorrect b) could be ok 4) H f o (As (g) at 25 o C ) = kj/mole a) obviously incorrect b) could be ok Lecture Notes C: Thermodynamics I(cont) Distributed on: Wednesday, January 19, 2005 Page 4 of 7

5 Hydrogen (perhaps produced by solar energy) would be an ideal alternative to fossil fuels, since it does not produce pollutants or green house gases when burned. The problem is that it is a gas, and hard to store and transport. What volume of hydrogen gas at 1.00 atm and 25 o C would be required to produce an amount of energy equivalent to that produced by the combustion of a gallon of octane (C 8 H 18 ) to form CO 2 (g) and H 2 O(l)? H f o (C 8 H 18 ) = kj/mol H f o (CO 2 (g)) = kj/mol H f o (H 2 O (l) ) = kj/mol Density of C 8 H 18 at 25 o C is g/ml Lecture Notes C: Thermodynamics I(cont) Distributed on: Wednesday, January 19, 2005 Page 5 of 7

6 The Bombardier Beetle defends itself by spraying nearly boiling water on its predators. It has two glands on the tip of its abdomen. Each gland has two compartments. The inner compartment holds an aqueous solution of hydroquinone and hydrogen peroxide. The outer compartment holds a mixture of enzymes that catalyze the following reaction: C 6 H 4 (OH) 2 (aq) + H 2 O 2 (aq) C 6 H 4 O 2 (aq) + 2 H 2 O (l) H = kJ hydroquinone hydrogen peroxide quinone When threatened, the beetle squeezes some fluid from the inner compartment into the outer compartment, and sprays the mixture (which is near its boiling point) onto the predator. As a model, suppose we mix a hydroquinone solution with a H 2 O 2 solution in the presence of the catalyst. What is the temperature of the solution after mixing 1.0ml of a 2.0M hydroquinone solution with 1.0ml of a 2.0M H 2 O 2 solution? Lecture Notes C: Thermodynamics I(cont) Distributed on: Wednesday, January 19, 2005 Page 6 of 7

7 Concept Suppose the concentration of H 2 O 2 is 2.5M, and that of hydroquinone is 2.0M. What happens to the final temperature of the solution? a) same as above b) higher than above c) lower than above Suppose the bug mixes 0.5ml of the hydroquinone solution with 0.5ml of H 2 O 2. What happens to the final temperature of the solution? a) same as above b) higher than above c) lower than above Lecture Notes C: Thermodynamics I(cont) Distributed on: Wednesday, January 19, 2005 Page 7 of 7