Lecture 40 - Topics for the day Administrative stuff...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Topics for the day Administrative stuff Homework problems Entropy as probability (continued) Second Law of Thermodynamics Third Law of Thermodynamics Administrative stuff Homework Quest homework just posted, due 10 am next Wed ALEKS objective due midnight Friday 5/7 (removed 4 topics - down to 11 topics) Practice exam coming, answer key available Monday I still owe you a practice exam grades on blackboard my apologies Homework problems The overall class average for HW5 is 70% Here are the questions that people found harder: 1 2 3
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
A. False B. True iClicker Time A closed system can exchange energy with the surroundings Open Closed Isolated Calculate Δ H˚ for the reaction 2N 2 (g) + 5O 2 (g) 2N 2 O 5 (g) given the following data: H 2 (g) + 1 / 2 O 2 (g) H 2 O(l) Δ H˚ = –285.2 kJ mol –1 N 2 O 5 (g) + H 2 O(l) 2HNO 3 (l) Δ H˚ = –76.6 kJ mol –1 1 / 2 N 2 (g) + 3 / 2 O 2 (g) + 1 / 2 H 2 (g) HNO 3 (l) Δ H˚ = –174.6 kJ mol –1 Calculation time (2) (3) (1) Balanced target eqn Add the Δ H values Reverse any reactions necessary Different stoichiometric coefficients? Add the modified equations Reverse reaction (2) and multiply by 2, to get N 2 O 5 on the product side: 4HNO 3 (l) 2N 2 O 5 (g) + 2H 2 O(l) Δ H = –2 × –76.6 kJ mol –1 = +153.2 kJ mol –1 (4) Multiply reaction (3) and Δ H by 4, to get the right amount of N 2 on the reactant side: 2N 2 (g) + 6O 2 (g) + 2H 2 (g) 4HNO 3 (l) Δ H = 4 × –174.6 kJ mol –1 = –698.4 kJ mol –1 (5) 4HNO 3 (l) 2N 2 O 5 (g) + 2H 2 O(l) Δ H = 153.2 kJ mol –1 (4) 2N 2 (g) + 6O 2 (g) + 2H 2 (g) 4HNO 3 (l) Δ H = –698.4 kJ mol –1 (5) 2N 2 (g) + 6O 2 (g) + 2H 2 (g) 2N 2 O 5 (g) + 2H 2 O(l) Δ H = –545.2 kJ mol –1 (6) 2N 2 (g) + 6O 2 (g) + 2H 2 (g) 2N 2 O 5 (g) + 2H 2 O(l) Δ H = –545.2 kJ mol –1 (6) Reverse reaction (1), and multiply by 2, to get 2H 2 O on the reactant side: 5 Compare the result to the target equation. 2N 2 (g) + 5O 2 (g) 2N 2 O 5 (g) We need to get rid of one O 2 and the H 2 from one side and the H 2 O from other side. 2H 2 O(l) 2H 2 (g) + O 2 (g) Δ H = –2 × –285.2 kJ mol –1 = + 570.4 kJ mol –1 (7) Add to equation (6): 2N 2 (g) + 6O 2 (g) + 2H 2 (g) 2N 2 O 5 (g) + 2H 2 O(l) Δ H = –545.2 kJ mol –1 (6) 2H 2 O(l) 2H 2 (g) + O 2 (g) Δ H = + 570.4 kJ mol –1 (7) 2N 2 (g) + 5O 2 (g) 2N 2 O 5 (g) Δ H = (–545.2 + 570.4) kJ mol –1 = 25.2 kJ mol –1 4 5 6
Background image of page 2
A. glass B. lead C. copper D. water iClicker Time Consider the following speciFc heats: copper 0.284 J g –1 ˚C –1 ; lead 0.159 J g –1 ˚C –1 ; water 4.18 J g –1 ˚C –1 ; glass, 0.502. Which substance, when warmed, would be more likely to maintain its heat and keep you warm through a
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

Lecture 40 - Topics for the day Administrative stuff...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online