# Questions on Chemistry

1. May 5, 2005

### courtrigrad

Hello

If we have a number of trials, and are given the concentration of 3 reactants is this how you would find the rate law? Also how would you find the average constant?

$$1 : 0.001 \ \ 0.1 \ \ 0.2 \ \ 1.2 \times 10^{-6}$$
$$2 : 0.001 \ \ 0.4 \ \ 0.2 \ \ 0.48\times 10^{-5}$$
$$3 : 0.003 \ \ 0.1 \ \ 0.2 \ \ 108\times 10^{-7}$$
$$4 : 0.003 \ \ 0.4 \ \ 0.4 \ \ 86.4 \times 10^{-6}$$

The 4 numbers are the # of trials, and the three numbers beside them are the concentration of the three reactants. The last number is the rate in $$mol/L*s$$

So using the rate law $$k[A]^{m}^{n}[C]^{z}$$ I know that in trial 1 the rate goes down by a factor of $$1/4$$. A and C are the same but B changes. Does that mean B's reaction order is 1/4?
Any help in trying to find the average value of the rate constant and the rate law would be appreciated.

Thanks

Last edited: May 5, 2005
2. May 5, 2005

### courtrigrad

would i have to compare rates from three experiments?

3. May 5, 2005

### bross7

You were close, just one manipulation, and one final step to get the answer.

When trying to find the rate law pick the two trials where the concentrations of two of the substances stay the same and only one changes then solve for the rate law by taking the ln of the solution.

$$\frac{rate2}{rate1}=\frac{k[A]^x ^y [C]^z}{k[A]^x ^y [C]^z}$$

Say we take runs 1 and 2 where the 1st (A) and 3rd (C) reactant concentrations stay the same so we can just cancel out the A's, C's and k's: the reaction constant is the same at constant temperature which we are assuming with the experimental data:

$$\frac{1.2\times{10^-6}}{0.48\times{10^-5}}=\frac{(0.1)^y}{(0.4)^y}$$

One of the rules of exponents allows us to simplify the right side:

$$0.25=(0.25)^y$$

Finally take the ln of both sides to bring down the unknown y and solve for the rate order value for the "B" concentration.

$$ln(0.25)=yln(0.25)$$
$$y = 1$$

4. May 5, 2005

### Artermis

I believe that because from Trial 1 to Trial 2, goes down 1/4th while A and C remain constant, and the rate also changes by a factor of 1/4th, this means that is zeroth order in relation to the rate constant, so I believe that you can strike it from your rate law; meaning that the rate law will now only depend on A and C.

If I am on the right track, let me advise you to use that knowledge to find the exponents for A and C. Hint: for [C] look at Trials 3 and 4 and apply your knowledge of the rate laws.

Correct me if I'm wrong, it's been a while since I did rate of reactions, so I may be a little fuzzy.

-Art

EDIT: Hrmph never mind then, I must be incorrect. bross, good solution

5. May 5, 2005

### courtrigrad

so would the reaction equation be $${k[A]^0 ^1 [C]^0}$$ making it an overall order of 1? How do you find k? k would be 1 because it cancels out and has no effect on the reaction?

Last edited: May 5, 2005
6. May 6, 2005

### bross7

To find the rate order of A and C you have to repeat the same process as you did for B using the rates where B and C stay the same, to find A, and A and B stay the same, to find C.

Then to find the rate constant k, just take any one of the three trials and sub the data into the equation you have found as the only unknown is k.

7. May 6, 2005

### Gokul43201

Staff Emeritus
You're right that you're wrong !

courtigrad : you have 4 equations in 4 unknowns, so you should be able to solve them.

There's a simple way, by inspection, that will work. You just have to choose the right lines to compare. From bross' work, you know that the rate is linear (first order) in ; meaning that when you double or triple , the rate will correspondingly double or triple.

Next compare lines 1 and 3. What concentrations change between these trials ? By what factor ? And how does this change the rate ? So, what can you tell from this ?

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