Aqa chem4 15th june 2011 (resit) thread

Just so you know, June 2009 Unit 4 is from the old specification, but is pretty similar to our current one. Here you go

hey i'm doing edexcel but i figured it doesnt make a difference with this question, i don't get rates

can someone explain why it's D please

June 2008 Q2 (a) Why is A 2nd order?

I'll try and explain this.
First assume that the value of

and [NO] are both 1.
From the description of the rate equation:
Rate=k

[NO]^2 = k[1][1]^2 =1 (NO is squared)
Now if you increase the conc of both H2 and NO by 3 times then:

Rate=k[3][3]^2 =3x9 =27

Hope that makes sense

Ok its easier if we look at order with respect to B.
In experiments 2 and 3 the I.concentration of A is the same (7.80x10^-2)
So any change in rate would mean its due to B.
And since the rate falls from (5x10^-5) to (2.50x10^-5) and the conc of B also decreases by half this must mean B has an order of 1.

Now with respect to A.
Look at experiments 1 and 2.
We already know that the order with respect to B is 1. So if the rate was to increase (for e.g) from 1 to 2 then this would mean that conc of B has been increased by the same amount (this is assuming A remains constant).
So when concentration of B was decreased from (9.60x10-2) in experiment 1 to (2.40x10^-2) in experiment 2, we would expect the rate to decrease by the same factor which is 4 but the rate remains constant. And we can see that the concentration of A was increased in experiment 2. Which must mean that A is having an effect on the rate of reaction. Now to work out the order you must see whats happening.

B is going down by a factor of 4 and A is going up by a factor of 2 (this change causes the rate to be the same at the end of both experiments).
So due to B the rate would go down from (5.00x10^-5) to (1.00x10^-5). But it stays constant as A is increased by a factor of 2. And since rate is increasing by a factor of 4 (even though A is only increased by a factor of 2) this must mean the order of A is 2.

2^2=4


Hope you can follow all that I've explained


Edit: ps sorry for the long explanation, but I think it will be worth it if you understand this as these questions can be quite tricky!!!

does anybody know how much detail we need to know about acid anhydrides?
Do we just have to know how to name, or know how to do equations and mechanisms etc. ?

When a 0.218 mol sample of hydrogen iodide was heated in a flask of volume V dm3, the following equilibrium was established at 700 K.
2HI(g) -----> H2(g) + I2(g)
The equilibrium mixture was found to contain 0.023 mol of hydrogen.
(i)Calculate the number of moles of iodine and the number of moles of hydrogen iodide in the equilibrium mixture.

Number of moles of iodine:

Number of moles of hydrogen iodide:

Can somebody please help me STEP BY STEP on how to do a question like this??? thank you in advance
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Need to learn chapter 11 today! :/

Can anyone explain why the hydrogen labelled 'd' proudes a doublet? Is it because it is attached to 2 singlets due to the 2 -COOH groups?

(CH3)2CH-CH(COOH)2

The hydrogen labelled 'd' is the one before the 2 -COOH groups.