I am trying to complete my lab and I have no idea how to do these back questions

Question

I am trying to complete my lab and I have no idea how to do these back questions. There are multiple pages of the same concept so if you could explain how to do this first set, I am capable of doing the rest on my own.

The lab was activation energy and effect of a catalysis.

In flask A i had 1.0mL of starch, 1.2 x10^-2 M 2.0 mL of Na2S2O3, .20M 4.0mL KI, .20M 4.0 mL of KNO3. In flask B i had .20M 2.0mL of (NH4)2S2O8 and .20M 6.0 mL of (NH4)2SO4. I mixed the two together until they turned blue.

Here is the data on how long it took the color change to happen

Now here are the questions I don't know how to solve with this data.

Please explain!!! I am so lost

determination 10 12 entration of reactants after ng 042M 042M 042M 042 M 021 M 021M 02M 2 M I, M tial time e of color change reaction time, s perature, °C -10 23 33 43°C ial time e of color change

Explanation / Answer

The first picture is cut on the left, so I can't read very well the data, but I do have an idea. So, I'm gonna give you some guidelines and tips for you to do this. It's not that hard to do it, you only need some guidance, and I will give that:

First, these kind of reactions is to determine the speed of reaction in order to calculate reaction rate (which is the speed) and constant rate (k). Now, depending on the order of reaction (zero, first, second order) the values you'll get will be different.

Now the rate of reaction would have to follow the next expression:

r = -dA / dt Where A is concentration of reactant and t is time.

You can also use this expression:

r = k[A]n

Where "n" is the order of reaction

For a zero order reaction, the equation to use is:

dA/dt = -kA

A = Ao - kt

For a first order reaction:

lnA = lnAo - kt

For a second order reaction:

1/A = 1/Ao + kt

According to your data, this is a first order reaction (I realize because your data is asking for a lnK).

Now, in addition to this, you need to calculate the activation energy, which you'll calculate following this expression:

k = Ae(-Ea/RT) Where A is a constant, k is the constant rate, R is the gas constant in Joules and Ea is activation energy.

In order to solve for Ea:

lnk = lnA - Ea/RT arranging the equation:

lnk = lnA - Ea/R (1/T) (2)

Now that we know this, let's try to solve your lab:

You already have the concentration of the reactant (You put that there as average) and you also for each measurement have the time, so to get the rate constant, just divide that value between the value of time to get the rate:

r = 2.55x10-6 / 192 = 1.33x10-8 M/s

To get the constant rate:

r = k[A] ---> k = r/[A]

You already calculated r, so solve for k.

Then calculate the lnk (kjust put that in your calculator)

The temperature in °C is the same that you have on your data, and to pass it to K, just sum 273:

K = -10 + 273 = 263 K

Then, calculate 1/T:

1/263 = 3.8x10-3 K-1

When you do that with the 4 measurements, you will have 4 values of lnK and 1/T, then plot lnK vs 1/T (With lnK in the y axis and 1/T in the x axis), doing this you'll have an expression similar to the one in (2) (See above).

From there, you'll have a linear equation and from there you can have the slope and y intercept:

lnk = lnA - Ea/R (1/T)

lnK = y

lnA = y intercept

Ea/R = slope

When you plot the data, the linear equation will give you the slope, and with that value solve for Ea using R = 8.3144 J/mol K

Finally 1/A you'll get it after you calculate y intercept, that's lnA. Solve for A (A = eyintercept) and then 1/A.

Hope this helps you and give you the start of this.

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