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Help!! Answer this question ??

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(edited 7 years ago)
Original post by iamspiderman
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helppppp
i know you have to find out the orders first but NO concentration, experiment no. 1,2,3 which are 0.001, 0.002 and 0.003 not sure which order this is ?? As the figures are not doubling ?


Can you type out the exact phrase of the question please?
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(edited 7 years ago)
Original post by iamspiderman
Using the results from the six experiments given, determine the values of m , n, k analyse the rates of reaction and also determine the overall reaction order (with respect to concentration of reactants )


m: In the table, look for two experiments you can compare, between which the concentration of NO doubles, but the concentration of H2 stays the same. Look at the rates of reaction for these two experiments. If the rate is the same, order is zero. If the rate doubles as concentration doubles, order is one. If the rate quadruples, then order is two.

n: Exactly the same here: Look for two experiments you can compare, between which the concentration of H2 doubles, but the concentration of NO stays the same.

Assume you understand the rest of the question seeing as it's the orders that you mentioned?

Hope that helps! :smile:
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(edited 7 years ago)
Original post by iamspiderman
Thank you! :smile: this does help


For NO
I chose 0.001 , 0.002 and 0.003
Whilst h2 1-3 concentration is the same
So the concentration is going up by 1 an the rates are 0.002, 0.008 and 0.018
I'm don't know which order this would be ? I don't think it's order 2 ....


Sorry, didn't see this! Just look at the difference in rate between 0.001 and 0.002. This shows a double in concentration. Now compare the change in rate between 0.002 and 0.008, which shows that as concentration doubles, rate quadruples (x4). This is 2nd order.

0= concentration doubles, rate stays the same.
1= concentration doubles, rate doubles.
2= concentration doubles, rate quadruples.

Because we are only looking at two different concentrations, you only need to compare two rates.

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