Chemistry A2: Urgent Help Needed!

you need experimental values....

I think its

Zero Order, No change in rate
1st Order, Rate doubles as conc doubles
2nd Order, Rate quadrupiles as conc. doubles

Please correct me if needed on this though.

I need to revise for a module containing this today....

To work out the order of a reaction you need to have data that shows changes in concentration of each of the reactants and how they affect the rates of reaction.

I only know the order to that reaction because I have it written down in my notes as an example.

http://www.ocr.org.uk/Data/publications/past_papers_2007_january/A_GCE_Chemistry_Salters_2849_01_January_2007_Question_Paper.pdf

Take for example that paper. Question 5 part C. That's the type of question you will get in an exam if they ask you to work out the order of a reaction. Have a look at it and see if you can do it based on what we've said. If you can't, then let us know where the problem lies and we'll all help.



Yeah that's correct.

I could only understand the 1st experiment, but then the concentrations keep on changing in the 2nd,3rd and 4th experiments that it gets me lost.

I think its

Zero Order, No change in rate
1st Order, Rate doubles as conc doubles
2nd Order, Rate quadrupiles as conc. doubles

Please correct me if needed on this though.

I need to revise for a module containing this today....

For that question you need to consider each one of the reactants in turn.

Start with [NO]

Going from experiment 2 - 3, the concentration of NO doubles. Then you look at the rate of reaction in the far column. The rate has quadrupled.

As Loz said previously if the concentration doubles and the rate quadruples, the order with respect to that reactant is 2nd order. Thus you can say that for NO it is 2.

Now looking at [CO]

Between the experiments 1 and 2 the concentration of CO doubles (all the other reactants stay the same) but the rate of reaction does not change. If it doesn't change, then the order with respect to CO is zero (As Loz said)

Finally looking at [O2]

Between experiments 3 and 4 they have doubled the concentration of O2. But if you look at the rate of reaction, nothing changes. Thus you can say that the order with respect to O2 is zero also.

Notice that they only change one concentration at a time, leaving everything else the same.

Thank you very much

Now all i need help on is the graph questions, and writing the rate equation for the reaction. Care to help?

From the calculations you know that the order with respect to NO is 2, and 0 for the other reactants. This means that you only need to include NO in the eqaution.

The general rate equation for a reaction: A + B = Products

is

Rate = k x [A]^m x ^n

Where m and n are the orders with respect to those reactants.

As previously said, two of the reactants have 0 order so don't get included in the equation. [NO] has an order of 2.

Have a go at writing the equation now that you know how they are constructed.

As for the graph. Since it comes before the part where you work out the order of the reaction, just draw a suitable curve. The gradient of the curve is the rate of reaction. The intitial rate of reaction will be the gradient of the curve at time = 0. For this particular question, draw a tangent to the curve at t = 0 and then explain that you need to measure the gradient of that tangent to get the intitial rate of reaction.