# Rate of reaction based on concentrations of reactants and products

Guys I'm a bit rusty with chemistry and I am preparing for mcat. I have three problems from the workbook I am using and I still dont see how they came up with the answers that they did. Can someone please help.

Here are the two questions.

1. Dinitrogen pentoxide decomposes according to the following reaction:

2 N2O5 -> 4 NO2 + O2

The rate of any reaction can be expressed in terms of the disappearance of reactants or the appearance of products. These expressions for the rate must be positive and equivalent. Considering this, write three expressions for the rate of the above reaction.

2. The activation energy, Ea, for a reaction is exactly 100kJ/mol. At what temperature must the reaction be run, for the rate to exactly double compared to the rate at 25 C?

3. The half-life for the reaction 2 N2O5 ? 4 NO2 + O2 is 2.4 hours at 30 C.

a) Starting with 10 g of N2O5, how many grams will remain after a period of 9.6 hours?

b) Starting with 100 g of N2O5, how many grams will remain after a period of 16 hours?

c) What period of time is required to reduce 5.0x1010 molecules of N2O5 to 1.0x10 8 molecules?

thanks

thanks for your help.

I have 2 more questions.

1. If we increase the temperature of a reaction by 10 oC, the rate will double. What is the value of the activation energy, Ea, that makes this statement exactly correct for a reaction run at 25 oC and 35 oC?

I'm not sure the reasoning for this, what I looked up seems a bit confusing. Can you explain it to me?

2. For the reaction A + B -> C the following data are obtained at 25 C:

Initial [A] Initial [B] Initial Rate

1.0 M 1.0 M 3x10-2

0.1 M 1.0 M 3x10-3

1.0 M 0.1 M 3x10-4

a) What is the rate law for the reaction?

for this I have k=[A][B]

b) What is the numerical value for the rate constant?

k=[A][B]: = 3.0X10^-2(mol/L.h)/ (1.0M)(1.0M)= 0.03 L/mol.h

this was data from experienment 1.

c) Speculate on a reasonable equation for the rate determining step.

According to my book, I dont understand how they came up with the answer.

d) When the first experiment is performed at 50 oC, the initial rate is 3x10-1. What is Ea for this reaction?

I dont know how to approach this.

Do you know any good general chemisty books I can study from. For whatever reason O chem was so much easier for me than gen chem. Thanks for all your help.

Hi Again

In Q1 we will assume the rate at the lower temperature is 1 unit. At the higher temp it will be 2. As Rate =k[reactants] in both cases (and reactant are the same in both), k will be proportional to 1 at 298K and proportional to 2 and at 308K. A plot of lnk vs 1/T will give a slope of -Ea/R.

ln1 = 0 for the lower temp and ln2 = 0.69 at the higher temperature.

The gradient (dy/dx) gives a value of -0.69/10 and this equal -Ea/R. Rearrange this Ea = 0.068 x 8.31 = J (seems a small value) (Never done this calculation before)

I think there is data missing in the new question 2.

What are the units of rate in Q2? I assume the middle column refers to the [B].

Please check the value for the rate when [A] =1 and [B]=0.1 - there is a 100 fold decrease in rate for a ten fold decrease in concentration - something wrong here.

It looks from your calculation in part b seems to include B in the expression as 1st order. If this is true, the the value in the table should be 3 x 10-3 for a concentration of B of 0.1M

Depending on the information in the table and based on your calculation, I suspect the rate law should be

rate =k[A][B]

If this is correct, the rate determining step must be a slow initial collision between A and B to form an intermediate which then decomposes rapidly to give C

As far as Ea calculation is concerned we need to calculate k for the higher temperature.

Taking natural log (ln) of both sides of the Arrhenius equation gives

lnk = lnA - Ea/RT

Plotting a graph of lnk (on y axis) vs 1/T (on x axis) (using 298K and 323K as the two temperatures).

The slope will have a gradient of -Ea/R ( and R is the gas constant)

thanks for your help, it looks like I did mistype up the 2nd question. But I figured it out . Here is my last question. Do you think you can help me one more time?

Q.- The thermal decomposition of acetaldehyde is given by the following reaction:

CH3CHO ->CH4 + CO

At 800 K, the following data are obtained:

Reactant concentration Rate of decomposition of CH3CHO

[CH3CHO] (M) -d[CH3CHO]/dt (Ms-1)

0.100 9.0 x 10-7

0.200 36 x 10-7

0.400 14.4 x 10-6

a) Write the rate equation for the reaction.

my answer to this is Rate=k[CH3CH0]

b) What is the order of the reaction?

my answer: 1st order

c) Calculate the rate constant for the reaction at 800K.

my answer: using data from experiment 1

rate= 9.0x10^-8 (ms-1)/0.100= 9.0x10^-6 ms-1.

d) Calculate the decomp. rate at 800K at the instant when [CH3CHO] = 0.250

my answer: k= 9.0x10^-6ms-1 (0.250)(CH3CHO)= 2.25X10^-6

I wanted to confirm is my answers were right or wrong.

Hi

I think all of the latest answers are wrong because you have made a mistake determining the reaction order. Look at data again. What happens to the rate of reaction in the second experiment compared to the first? If the reaction is first order, the rate should double. However, in this example, there is a four fold increase in rate - so this is 2nd order kinetics

Rate proportional [reactant]^2 and rate = k[reactant]^2

This is confirmed when expts 2 and 3 are compared. You will need to recalculate the rest of the info on this basis

I am still having trouble trying to figure out the last two questions.

c) Calculate the rate constant for the reaction at 800K.

d) Calculate the decomp. rate at 800K at the instant when [CH3CHO] = 0.250

thanks for all your help. God bless you.

Sorry for the delay in replying but I have been away to visit relatives.

Reaction rate = k[CH3CHO]^2 so the order is Second order

k = Reaction rate/[CH3CHO]^2

= 9.0 x 10-7 Ms-1/(0.1)^2 M^2

= 9.0 x 10-7 Ms-1/0.01 M^2

= 9.0 x 10-5 s-1 M-1

Now you have the rate constant at this temperature, just put the values into the rate equation

Reaction rate = k[CH3CHO]^2

Reaction rate = 9.0 x 10-5 s-1M-1 x 0.25^2 M^2

= 5.625 x 10-6 M s-1

Hi

I retired from teaching some years ago but I will help where I can

The rate of reaction can be expressed as the rate of change in concentration of a substance.

A ---> B

Concentration of A ( written as [A]) is decreasing while is increasing

Rate of the reactioncan be expressed as the change (increase) in concentration of product/time

While the same rate can be expressed in terms of -(decrease in concentration of reactant/time

Notice the negative sign is used as the value for the concentration of reactants will be decreasing. This will make the rate positive whichever is used.

You can use calculus to show rate (I'm not so good at this)

Rate = - decrease in concentration of N2O5/time (or in calculus (-d[N2O5]/dt ) where d = delta and meaning the change in [N2O5] / time interval)

Rate = increase in concentration of NO2/time (or in calculus (d[NO2]/dt )

Rate = increase in concentration of O2/time (or in calculus (d[O2]/dt )

The second question is more difficult to me but involves the Arrhenius equation

where A = frequency factor (usually constant over a small temperature change, e = mathematical number relating to natural logs (ln), Ea is the activation energy, R is the gas constant and T is the temperature in K

A is a constant so we will calculate the value of e ^-Ea/RT for the lower temperature. As the gas constant R is quoted as 8.31 J K-1 mol -1 we will express Ea in Joules (= 100 000J)

-100000/8.31*298 = -40.382. So e ^-Ea/RT = 2.901*10^ -18

You will probably need to then rearrange this expression in term of T to find when the rate become 2*(2.901*10^ -18) ie 5.81*10^ -18. I can't remember how to do this

5.81*10^ -18 = e^(100 000/(8.31 * T))

Just by inspection, if I put a temperature of 308 into the expression, we get the answer 1.076*10^ -17 which is more than doubling the rate. Trying different values it appears that a temperature of 303.245K is about right.

I'm not sure how to do the rearrangement of the equation. Inspection value suggest 303.245-298 = 7.702oC

I'll send this but apologise for not being able to complete it. If you don't get the email, I'll post it on the website.

As for the other question, I've only come across t1/2 in radioactivity. If it means the same it will be the time taken for 1/2 of the substance to react.

On this basis in part a) 9.6 hours is 4 * t1/2 (four half life periods) or (1/2)^4 or 1/16. So there will be 1/16 of 10g left = 0.625g left.

In part b) 16 hours is 16/2.4= 6.67 half life periods. So the fraction of the 100g remaining will be 100*((1/2)^6.67) = 100 * 0.0098 = 0.982g left.

Hope this helps

I have spent ages with the maths and the calculator