# chemistry

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disappearance of NO in ...- Help!!!?

The following data were collected for the rate of disappearance of NO in the reaction 2 NO(g) + O2(g) ¨ 2 NO2(g).

Run: [NO](M) [O2](M) Initail Rate (M/s)
1 .0126 .0125 1.41 X 10^-2
2 .0252 .0125 5.64 X 10^-2
3 .0252 .0250 1.13 X 10^-1

(a) What is the rate law for the reaction?
(b) What are the units of the rate constant?
(c) What is the average value of the rate constant calculated from the three data sets?
(d) What is the rate of disappearance of NO when [NO] = 0.0750 M and [O2] = 0.0100 M?

(e) What is the rate of disappearance of O2 at the concentrations given in part (d)?

• chemistry - ,

Instead of me doing all of the typing why don't you show what you know how to do and explain clearly what you don't understand about the other parts. I can help you through the tough spots.
By the way, you can go to this site and read units for orders 1,2,and 3.
http://en.wikipedia.org/wiki/Reaction_rate_constant

• chemistry - ,

i don't know how to do any of it

• chemistry - ,

Look at trial 1 versus 2. O2 is same concn; NO is doubled. What happens to the rate? It is 4x (0.0141 x 4 = 0.0564 so the reaction is second order with respect to NO.
Now compare trial 2 versus trial 3. (NO) is constant while O2 changes by a factor of 2 (0.125 to 0.250). How much did the rate change? Looks like it doubled (0.0564 x 2 = 0.1128 or 0.113)? So double O2 and we double rate which makes O2 first order.
rate = k(NO)2(O2)1 and you really don't need the 1.
The site I left last time will answer b for you.
For c, use the rate law you have from part a and substitute the data in the three trials. Solve for k and take the average.

• CHEMISTRAH - ,

a) NO doubles = rate quadruples
NO is 2nd rate (2^2 = 4)
O2 doubles = rate doubles
O2 is 1st rate (2^1 = 2)

b) M^-2s^-1

c) 1 = 7105 M-2s-1
2 = 7105 M-2s-1
3 = 7118 M-2s-1
avg = 7109 = 7110 M-2s-1

d) rate = 7110(0.0750)^2(0.0100)
rate = 0.400 M/s

e) 1/2 rate of disappearance of NO
0.400 / 2 = 0.200 M/s

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