How do I find the rate equations of each of these reactions with the information given? Also, the overall reaction orders should be the exponents in the equations added together right?

If I have:
Initial Concentrations of X: Initial Concentrations of Y:
0.20 | 0.10
0.40 | 0.10
0.20 | 0.30

And the Initial Rates:
0.025
0.050
0.075

Thank you!

Question

How do I find the rate equations of each of these reactions with the information given? Also, the overall reaction orders should be the exponents in the equations added together right? 

If I have: 
Initial Concentrations of X: Initial Concentrations of Y: 
0.20 | 0.10 
0.40 | 0.10 
0.20 | 0.30 

And the Initial Rates: 
0.025 
0.050 
0.075 

Thank you!

anonymous · Answer

I'm confused as to what needs to go in R=k[x][y]

aaronq · Answer

First, to determine the rate law, you need to compare 2 trials where the concentration of one of the reactants is held constant and the other is changing.

For example, 1 and 2
 X         Y       rate
0.20 | 0.10   0.025 
0.40 | 0.10   0.050

[Y] is constant, but [X] is doubled, and the rate also doubled.

this means that [X] is to the power of 1.

Hypothetically, if the rate had changed from 0.025 to 0.1, then you would know it was $[X]^2$, because the rate would have quadrupled.

Do the same for Y.

anonymous · Answer

ok… so. R=k[X][Y]
If 1 is the exponent, you don't actually include it write? Or am I totally confused
@aaronq

aaronq · Answer

it's implied, you dont have to write it, but you could.

anonymous · Answer

Ok. Thanks. I had two lessons on this and neither were anywhere close to this example. 
For 'experiments' 2 and 3… Y tripled so what does that mean in terms of the equation:
R=k[X][Y]

aaronq · Answer

you have to look at the rates, how did the rates change from trial 1 to trial 3?

anonymous · Answer

tripled

aaronq · Answer

so the exponent is 1, right?