Question

I'm stuck on this homework and don't know where to start!

A student is trying to determine the rate law for a reaction involving 2 reactants, A, B and C using the Method of Initial Rates. The following initial rate data were obtained at 25 degrees C. (Note: the overall stoichiometry of the reaction is: A + B + 2C -> 3 D)

Experiment______Initial [A]______Initial [B]______Initial [C]_______Initial Rate of Reaction

1______________ 0.1 M_________ 0.1 M_________ 0.2 M _________4x10^-4 M/min

2_______________ 0.3 M ________0.2 M _________0.2 M __________1.2x10^-3 M/min

3 ______________0.1 M_________ 0.3 M _________0.2 M ___________4x10^-4 M/min

4______________ 0.3 M_________ 0,4 M _________0.6 M __________3.6x106-3 M/min

a) Write the rate law expression for the reaction

b) What is a plausible rate-determining step in the mechanism of this reaction?

c) What is the numerical value of k, the rate constant for this reaction?

d) In Experiment 1, what would the rate of the reaction be when exactly 1/2 of the A present initially has been consumed?

e) In Experiment 2, what is the maximum concentration obtainable for D? (Assume that the volume remains constant)

f) In Experiment 4, when the reaction has proceeded to completion, which reactant or reactants if any, and in what concentration, remain unconsumed?

Answer 1

The rate law expresess the relation between speed of reaction with its "k" constant and the concentration of the reactants raised to x number.

For a reaction like this:

Important: The x and y numbers are not equal to te coefficients of the reactants in the equation. These numbers are determined by an experiment.

The first thing to do is to write the rate law for this equation

Theoretically it is like this

Now we have to look at the experimental values.

When you are looking for experiments like this you need to identify if there´s a reactant that does not affect the rate of the reaction.

In this case if we take a look to experiment 1 and 3 we can see how the reactant B behaves:

The rate is not affected when the concentration of B is raised while the other 2 remain constant so we can say that the speed reaction is not affected by the concentration of B.

so

Now let´s take another look to the experiments, let´s see what happens when the concentration of A changes and C reamain constant.

Experiments 1 and 2, and experiments 2 and 3 can help us:

We see that when we multiply the concentration by 3 the rate of reaction is multiplied by 3:

So we can say that the concentration of A affects directly and in a linear way the rate of reaction, until now we have:

Now let´s take another look to the experiments, let´s see what happens when the concentration of C changes and A reamain constant. Experiments 2 and 4 can help us with that.

We can see that when we multiply by 3 the concentration of A the rate of reaction is also multiplied by 3

so in the equation the concentration of C is raided by 1:

The rate law expression for this reaction is:

.

The order of reaction is 2. A and C are raised to 1 each so and b is raised to 0. so 1 +1 +0 = 2

Now to get the k value we only have to substitue the values from any experiment, we get for experiment 1:

k = 0.02 M^-1 s^-1

If 1/2 of A is consumed for stoichiometry 2 moles of B are consumed for every consumed mole of A

The new concentrations will be

For A = 0.1/2 = 0.05 M

For C = 0.2 - (0.05*2) = 0.1 M

In experiment 2 we can see that for stoichiometry the limiting reactant is C.

Let´s assume we have 1 liter of solution

We have 0.3 moles of A, for every mole of A there should b 2 moles of C and 1 moles of B. Theoretically we should have 0.3 moles of B and 0.6 moles of C.

B and C are limiting A because we have less ammount that the theoretically needed.

so we see that we have an excess of A (will remain unchanged)

We also have 0.2 moles of B, for every consumed mole of B we need 2 of C. Theoretically we should have 0.4 moles of C, we only have 0.2 moles.

So C is the limiting reactant because is the one with the smallest ammount of moles.

So the maximun concentration of D that we could get is whey all C is consumed if we apply stoichiometry:

2 moles of C = 3 moles of D

since we assume to have 1 liter the maxium concentration of D is 0.3 M

In experiment 4

We have to do the same analysis we did for experiment 2.

If we do it we can see that there are 0.4 moles for B and theoretically we would need 0.4 moles of A and 0.8 moles of C. Since we have less for those laste 2 we can say that C is in excess.

A and B are in balance.

When reaction has been completed, A and C are 0M. We will only have C

1 moles of A needs 1 moles of B to react

we have 0.3 moles of A and 0.4 moles of B

Since is a 1:1 reaction, the remaining quantity of B is = 0.4 - 0.3 = 0.1 moles or 0.1M of B