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On the Spring 2010 exam, for number 17, it says to redo experiment 3 at 16.31 degrees, the rate is 7.53x10-4. How do you calculate the new k for this problem. The answer is 7.17x10-5

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The new k is calculated from the concentrations used in experiment 3 and the new rate. When the temperature of a reaction changes, the mechanism doesn't change (over small temperature changes) so the rate law expression is the same, you can plug everything in and use the same orders you determined at the original temperature.

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I'm having trouble with #14 on the Spring 2008 exam. How do you solve it?

14. You have found the following value in a table of equilibrium constants at 25ºC:

Cu2+(aq) + 4 NH3(aq) 􀀧 [Cu(NH3)4]2+(aq) Kc = 1.7x10-13

What is the equilibrium constant for the reaction:

2 [Cu(NH3)4]2+(aq) 􀀧 2 Cu2+(aq) + 8 NH3(aq)

----Answer----

This is similar to the lead bromide question... To get from the first equilibrium reaction to the second, we have to: 1)reverse the reaction; 2)multiply by 2. To convert the equilibrium constant, that means we have to: 1)take the inverse; 2)raise it to the 2nd power. So:

K{new} = ( 1 / (1.7E-13))^2 = 3.46E25

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17. You have found the following value in a table of equilibrium constants at 25ºC:

PbBr2(s) 􀀧 Pb+2(aq) + 2 Br-(aq) Kc = 6.29x10-6

What is the equilibrium constant for the reaction:

½ Pb+2(aq) + Br-(aq) 􀀧 ½ PbBr2(s)

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To get the second reaction, the first reaction must be: 1)reversed; 2)multiplied by 0.5. Therefore, the equilibrium constant must be: 1)inverted; 2)raised to the 0.5 power. So the "new" equilibrium constant value is:

( 1 / 6.29E-6 )^0.5 = 399

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On the exam -summer 2007 (exam 2) #9, I wasn't sure how you got the second grams (16.246 g / 331.208) because in the periodic table it said pb is 207.2, so what do I have to do to get it to 331.208?

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The source of the lead(II) is lead(II) nitrate. When you're looking at equilibrium problems (or kinetics problems, or any stoichiometry problems) you can usually just use the net ionic equation, but you need to include spectator ions when you're weighing out reactants.

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On exam - fall 2004 (Exam 1), I wasn't sure how to calculate #12. For #14 I converted it to grams then to the concentration but I don't seem to be getting the right answer either.

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That's an old exam... #12 has an error, so there is not correct answer listed, but here's how to do it. The rate of oxygen consumption is:

0.433mols/8.5oL/min = 0.0509 M/min oxygen consumed

From the balanced chemical equation, for every 5 mols of oxygen consumed in the reaction 4 moles of ammonia is consumed, so the rate of ammonia consumption is:

(0.0509 M/min oxygen)(4 moles ammonia / 5 moles oxygen) = 0.0408M/min