Desperately need Chemistry help quick

Question

2 NO(g) Cl2(g) <--> 2 NOCl(g)

The following data were collected for the reaction above. All of the measurements were taken at a temperature of 263 K.

http://i218.photobucket.com/albums/cc123/2cool4wrds/quiz.jpg

(a) Write the expression for the rate law for the reaction above.

(b) Calculate the value of the rate constant for the above reaction, and specify the units.

(c) What is the initial rate of appearance of NOCl in experiment 2?

(d) What is the initial rate of disappearance of Cl2 in experiment 4?

(e) Each of the experimental trials took place in a closed container. Explain or calculate each of the following:

(i) What was the partial pressure due to NO(g) at the start of experiment I?

(ii) What was the total pressure at the start of experiment I? Assume that no NOCl is present.

Answer 1

(a) Write the expression for the rate law for the reaction above

from 1 --&gt;2 they doubled Cl2 ,...&amp; it caused the rate to double, so it is first order with respect to Cl2

from 1 --&gt;3 , they doubled the NO, which caused the rate to go 4 times faster, so it is second order with respect to NO

your answer is: Rate = k [NO]^2 [Cl2]

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(b) Calculate the value of the rate constant for the above reaction, and specify the units.

Rate = k [NO]^2 [Cl2]

k = Rate /  [NO]^2 [Cl2]

k = (0.60 mol/litre / min)  /  [ 0.15mol/litre]^2 [0.15 mol/litre]

k =  (0.60 mol/litre / min)  /  (0.003375 mol3 / L3 )

k = 177.77

your answer is(2 sigfigs):  k = 1.8 e2 L2/mol2 -min

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(c) What is the initial rate of appearance of NOCl in experiment 2?

I am assuming that the photobucket data table meant to say that the rate given was the rate of disappearance of &quot;Cl2&quot;

since 1 Cl2(g) --&gt; 2 NOCl(g),... NOCl appears twice as fast as Cl2 disarrears, so if Cl2 disappearance rate is 1.2 M/min,...

your answer is NOCl appears at the rate of 2.4 Molar/min

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(d) What is the initial rate of disappearance of Cl2 in experiment 4?

since the rate law is:

Rate = k [NO]^2 [Cl2]

Rate = (1.7777 e2 L2/mol2 -min)  [0.25]^2 [C0.25]

rate = 2.78 M/min

your answer is(2 sigfigs): 2.8 M/min

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(e) Each of the experimental trials took place in a closed container. Explain or calculate each of the following:

(i) What was the partial pressure due to NO(g) at the start of experiment I?

PV=nRT  , using the given ratio of 0.15 moles in 1 Litre @ 263K

P(1 Litre) = 0.15 moles((0.08206)(263K)

P= 3.237 atm

Your answer is(2 sigfigs): 3.2 atm  

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(ii) What was the total pressure at the start of experiment I? Assume that no NOCl is present.

since it was also 0.15 molar Cl2 ,,, its partial pressure is also 3.237 atm

the two total 6.474 atm

your answer is(2 sigfigs) : 6.5atm