Question

I have this question half way done I just need it checked to make sure I'm doing it correctly

Answer 1

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

Q + X products

Trial [Q] [X] Rate
1 0.12 M 0.10 M 1.5 × 10-3 M/min
2 0.24 M 0.10 M 3.0 × 10-3 M/min
3 0.12 M 0.20 M 12.0 × 10-3 M/min

Answer 2

3.0-2(m/min)/1.5-2(m/min)
(0.24/0.12)^x(0.1/0.1)
2^x=1
2=x

Answer 3

@kirbykirby from what I remember you were really good at explaining chem so can you help?

Answer 4

\( \large \text{rate}=k[Q]^x[X]^y\)

Now choose 2 experiments where the initial concentrations are constant.
\[ \large \frac{\text{rate (2)}}{\text{rate (1)}}=\frac{k(0.24 M)^x(0.10M)^y}{k(0.12 M)^x(0.10M)^y}=\frac{3.0 × 10^{-3} M/min}{1.5 × 10^{-3} M/min} \\ \, \\
2^x=2\\
x=1\]

Now to find \(y\), you know x, so to find y you can choose any rates to compare, so say rate (3) and rate(1). :

\[ \large \frac{\text{rate (3)}}{\text{rate (1)}}=\frac{k(0.12 M)^1(0.20M)^y}{k(0.12 M)^1(0.10M)^y}=\frac{12.0 × 10^{-3} M/min}{1.5 × 10^{-3} M/min} \\ \, \\
\large 2^y=8=2^3\\
y=3\]

\(\therefore \text{rate law} =k[Q][X]^3\)

Then to find the rate constant, you can rearrange the above equation and say look at trial 1: \[\large k = \frac{\text{rate}}{[Q][X]^3}= \frac{1.5 × 10^{-3} M/min}{(0.12 M)(0.10 M)^3}=12.5\,min^{-1}M^{-3}\]

Answer 5

Would I just do 12.0-2(m/min)/3.0-2(m/min)
(0.12/0.24)^x(.20/.10)
1/2=2

Answer 6

Wait why would the first one be 2 instead of 1 like I had it

Answer 7

So the overall reaction would be (12.5min^-1m^-3)xy^3

Answer 8

\[\large \frac{\text{rate (2)}}{\text{rate (1)}}=\frac{\cancel{k}(0.24 M)^x\cancel{(0.10M)^y}}{\cancel{k}(0.12 M)^x\cancel{(0.10M)^y}}=\frac{3.0 ×\cancel{ 10^{-3} M/min}}{1.5 × \cancel{10^{-3} M/min}} \\ \, \\
\large \frac{0.24^x}{0.12^x}=\frac{3.0}{1.5}\\ \, \\
\large 2^x=2 \\
\large x=1\]

Answer 9

yes

Answer 10

Thank you ^-^

Answer 11

Do you think you could answer one more question?

Answer 12

maybe, depends if I know it or not :)

Answer 13

Coo<3

A catalyst increases the rate of a reaction by

providing an alternative reaction pathway with lower enthalpy change.

lowering the potential energy of the reactants in the reaction.

providing an alternative reaction pathway with lower activation energy.

lowering the potential energy of the products in the reaction.

Answer 14

I'm pretty sure it's b but I just wanted to check

Answer 15

C, it lowers the activation energy

Answer 16

the potential energies remain the same

Answer 17

Okay thank you <3

Answer 18

yw :)