[mass]

Hi im doing aqa chemistry and im in year 14. Rates of reaction is causing me a lot of trouble, i cant get my head around it, if someone can help i would be grateful.

heres 1 qu

Heres a question:

Exp A B RATE
1 1 1 100
2 2 1 200
3 1.5 3 1350


if anyone could explain how to find order for a or b i wud be grateful

[sdekivit]

here's how you could approach it:

first look at reagent A, while keeping the other constant: when we double the cncentration, the reaction rate doubles. This means that s ~ [A]

same for reagent B only here it gets a bit complicated:

increasing three times increases the rate 13,5 times but [A] also increased 1,5 times cause s to increase 1.5 times ( = 150 what btw?)

--> thus increasing increases s a factor 9 and thus s ~ ² (when in creases 3 times, s increases 3^2 times)

this yields the equation s = k * [A] * ²

Next time you must post what you've tried before, but this was an attempt to try explain it to you.

[english]

You have:

Exp.       [A]          Rate
1                 1       1       100
 
2                 2       1       200

3                 1.5    3       1350


In experiment 1 and 2 is constant .

Solve for [A]:

2^x/1^x = 200/100
2^x=2
x = 1

That's one way to solve for [A] that involves arithmetic and is easier to see.