At 700 K acetaldehyde decomposes in the gas phase to methane and carbon monoxide

Question

At 700 K acetaldehyde decomposes in the gas phase to methane and carbon monoxide. The reaction is:
CH3CHO(g)CH4(g)+CO(g)
A sample of CH3CHO is heated to 700 K and the pressure is measured as 0.47 atm before any reaction takes place. The kinetics of the reaction are then followed by measurements of total pressure and these data are obtained:

a. Find the rate law.

4. Rate=PCH3CHOt=4.5×104atm1s1PCH3CHO2

b. Find total pressure after 1.41×104 s .

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

To know this, you need to plot the data and determine the order of reaction.

For a first order reaction: Ln P = LnPo - kt

For a second order reaction: 1/P = 1/Po + kt

As we have a total pressure with t, we need to know the individual pressures of each compound:

CH3CHO ---------> CH4 + CO

i. 0.47 0 0

e. 0.47-x x x

Pt = PCH3CHO + PCH4 + PCO = 0.47-x + x + x = 0.47+x --> x = Pt - 0.47

For t = 0 x = 0

P(0) = 0.47 atm

p(1000) = 0.55 - 0.47 = 0.08 ---> PCH3CHO = 0.47-0.08 = 0.39 atm

P(3000) = 0.65 - 0.47 = 0.18 ----> PCH3CHO = 0.47-0.18 = 0.29 atm

P(7000) = 0.75 - 0.47 = 0.28 ----> PCH3CHO = 0.47-0.28 = 0.19 atm

Now, we can plot this data into a first order reaction (plot lnP vs t). The slope, intercept and run is:

slope = -1.27x10-4

intercept = -0.800031

r2 = 0.98661

It could be a first order reaction, however, let's plot a second order reaction (1/P vs t). If r is closest to 1 then we'll go with this:

slope = 4.48x10-4

intercept = 2.1176

r2 = 0.99994

This means that this is a second order reaction, and the rate law is the 4th option (see the value of slope, that is similar to the value in that option).

Now for the pressure:

1/P = 1/0.47 + 4.48x10-4 (1.41x104)

1/P = 8.4344

P = 0.1186 atm

Hope this helps

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