----Question----

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

----Answer----

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.

----Question----

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