Question

PLEASE HELP!
Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided.

X + Y products


Trial [X] [Y] Rate
1 0.20 M 0.15 M 2.4 × 10-2 M/min
2 0.20 M 0.30 M 4.8 × 10-2 M/min
3 0.40 M 0.30 M 19.2 × 10-2 M/min

Answer 1

select 2 trials where the concentration of one of the reactants remains constant.
From there, you can determine the exponent for that reactant based on the change in the rate.

Answer 2

did I do it correctly?

Rate= k [X]^x [Y]^y
2.4 E-2 M/min = K*.2 moles/liter*(.15moles/liter)^3
2.4 E-2 M/min=k*6.75 E-4 moles^4/^4
k=35.6 l^4/min*moles^3

Answer 3

which 2 trials are you using?

Answer 4

whoops rate = k [X][Y]^3
2 & 3

Answer 5

you would use trial 2 &3 to find the exponent of X because Y is constant.

2 0.20 M 0.30 M 4.8 × 10-2 M/min
3 0.40 M 0.30 M 19.2 × 10-2 M/min

\(rate=[X]^n\)

rate increased by 19.2/4.8= 4, while the concentration doubled

\(1=[1]^n\)
\(4=[2]^n\) n=2

thus \(rate=k[X]^2[Y]^m\)

now use 1 & 3 to find the exponent of Y