Help me solve this!!

Question

The following are experiments performed at 28.36°C:

Exp’t # [feldomite]initial [anicium]initial [dolomide]initial Rate initial (M/second)

1 0.963 0.224 0.422 1.295 x 10-3

2 0.321 0.224 0.422 4.317 x 10-4

3 0.321 0.112 0.422 4.316 x 10-4

4 0.642 0.448 0.211 4.318 x 10-4

5 0.642 0.448 0.633 1.297 x 10-3

1.) What is the correct rate law expression for this reaction including a value of “k” with correct units?

2.) If Experiment #4 is repeated at 15.67°C, the initial rate is 1.839 x 10-5 M/second. What is the activation energy for this reaction?

Answer 1

No matter what the order of the reactants, the rate law is going to look something like this:

rate = k * [feldomite]^m * [anicium]^n * [dolomide]^p

Here, k is the rate constant, and m, n, and p are the orders of the reactants. Our goal is to figure out m, n, and p.

If you look at the initial rates data you are given, a lot of the entries in that table are the same! This is intentional, so that if you write out the ratio of two rates, the concentrations will cancel out, allowing us to solve for the remaining terms. Let's see how that works. Let's write out the ratio of experiment 1 and 2.

( 1.295 x 10-3 / 4.317 x 10-4 ) = ( ( k * 0.963^m * 0.224^n * 0.422^p ) / ( k * 0.321^m * 0.224^n * 0.422^p) )

Let's go ahead and simplify the fraction on the left, and cancel out what we can on the right. We don't know k, but we know it is the same as long as we are at the same temperature, so we can cancel that out too.

3 = ( 0.963^m ) / ( 0.321^m )

3 = ( 0.963 / 0.321 )^m

3 = 3^m

So, 3 to what power equals 3? Looks like m=1. We can do the same thing to figure out n and p using the other rows of the table. rate2/rate3 simplifies to:

1 = 0.224^n / 0.112^n

1 = ( 0.224 / 0.112 )^n

1 = ( 1/2 )^n

1 is 1/2 to what power? The 0 power (anything raised to the zero power is 1).

rate4/rate5 simplifies to:

0.333 = ( 0.333 )^p

So p=1. Putting together what we have so far, our rate law is:

rate = k * [feldomite]^1 * [anicium]^0 * [dolomide]^1

rate = k * [feldomite] * [dolomide]

What is k though, and what units should it have? Let's start with the units. The table tells us the units for the concentrations and the rates, so let's plug them into the equation and solve for k's units

M/s = k * M * M

M/(s*M*M) = k

1/(s*M) = k

k = s^-1 M^-1

Now that we have our units, let's plug in some numbers and see what value we get. I'm using line 1 from the table, but any of them would give you the same k!

1.295 x 10-3 = k * 0.963 * 0.422

k = 0.003187 = 3.187 * 10^-3 M^-1 s^-1

There's our answer to part 1. For part 2, we need the Ahrrenius equation.

k = A * e^( -Ea / (R * T) )

Here, k is the rate constant, A is another constant called the Ahrrenius pre-exponential factor, Ea is the activation energy for the reaction, R is the ideal gas constant (you've probably seen this guy before), and T is the temperature. (in Kelvin!) Great, we've defined everything! Now let's look at a different form of the eqation...

To get there, take the natural log of both sides of the equation.

ln(k) = ln (A * e^( -Ea / (R * T) ) )

Split up the logarithm using log(A*B) = log(A) + log(B)

ln(k) = ln(A) + ln(e^(-Ea/(RT))

And simplify that second term using ln(e^x) = x

ln(k) = ln(A) - Ea/RT

This is the form I want to talk about! This is the linear form. That is to say, it "looks like" y = m*x + b. The y is the ln(k) and the intercept, b, is ln(A). x is 1/T and the slope, m, is -Ea/R. So what this equation gives us is a linear relationship between ln(k) and 1/T, meaning it tells us about the temperature dependence of the rate constant.

If we want to know the activation energy, we just need the slope. Slope = rise over run = change in y over change in x = change in ln(k) over change in 1/T.

m = ( ln(4.318 x 10-4) - ln(1.839 x 10-5) ) / ( 1/301.36 - 1/288.67 )

m = 3.156 / -0.000146

m = -21636.4

m = -Ea/R

-21636.4 = -Ea/ 8.314

Ea = 179885 Joules/mole = 180. kJ/mol

The units for Ea come from the units for R. I used R = 8.314 Joules / (mole K)

Answer 2

Use functional proportionality kC_A^aC_F^fC_D^d = r therefore you can use data that has same initial conditions to set up proportions. For the following, the subscript stands for the particular run:

For F: ( C₂/C₁)^f = r₂/r₁ or (.321/.963)^f = .4317/1.295 or .333^f = .333 (you can usually tell by inspection what the exponent is... f =1 Formally f = ln(r₂/r₁)/ln(C₂/C₁)

For A do the same thing for runs 2 and 3

For D do the same thing for runs 4 and 5

Then take any run and plug in all the exponents, concentrations, and r and solve for k:

kC_A^aC_F^fC_D^d = r

The k units will cancel the concentrations and introduce concentration/s. eg 3rd order total (a+f+g =3) would have a k with units of (liters/mole)²(1/s)

The two-point relation for k changing with temperature assumes k = Ae^-E/RT. With concentrations held constant and taking the ln of the r ratio (=k ratio)

ln(r₂/r₁) = -(E/R)(1/T₂-1/T₁) Remember that R = 8.314 J/mole K, T is in K and E will be in Joules

Rearranging E = -Rln(r₂/r₁)/(1/T₂-1/T₁)

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