7. Problem 34 (pg 603) Consider the reaction: 2 H2O3 (aq)- 8 H20(I) + O2(g) The

Question

7. Problem 34 (pg 603) Consider the reaction: 2 H2O3 (aq)- 8 H20(I) + O2(g) The graph below shows the concentration of H202 as a function of time 1.2 1.0 e 0.8 0.6 0.4 0.2 HO2 0 10 20 30 4050 60 70 80 Time (s) Aen Miro M220s Use the graph to calculate each quantity a. The average rate of the reaction between 10-20 s =-1. 2. 2 20-lo b. The instantaneous rate of the reaction at 30s c. The instantaneous rate of formation of O, at 50 s d. If the initial volume of the H O, is 1.5 L, what total amount of O, (in moles) is formed i the first 50 s of reaction? -17 H202

Explanation / Answer

You have multiple questions in one single post. Please post your questions per separate so you can be answered better and faster and due to guidelines we are not allowed to answer multiple questions in a same post.

I will help you for now in problem 9 and 10.

For problem 9, the best way to determine the order, is the way you are doing that. However, there is just one value there that is actually out of focus. This value is 0.0016 when it should be 0.016. Why I said this? because if you took the first and the last rate, you'll get a smooth value of n. and the first value and the third value are really close and have a relation one of another. the second is 10 times smaller thant the actual value that it should be. So ignoring the fact that this value may be wrong:

A2/A1)n = r2/r1

(0.30/0.15)n = 0.016/0.008

2n = 2 ---> n = 1

For question 10, plot the data for a zero, first and second order. The one closest to 1 (in r2 value) will be the one that the order belongs.

for a zero order reaction, plot C vs t, and the data obtained is:

r2 = 0.95803

y = 0.907 - 0.00386x

For a first order reaction, plot lnC vs t, and the data obtained is:

r2 = 0.999999999

y = -0.00023 - 0.007804x

Finally for a second order reaction, plot 1/C vs t

r2 = 0.954641

y = 0.6359799 + 0.0183448x

So, this is a first order reaction.

Concentration at t = 250 s

lnC = -0.00023 - 0.007804(250)

lnC = -1,95

C = 0.1421 M

Hope this helps.

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