Question

Acetaldehyde decomposes to methane and carbon monoxide according to the following balanced equation:

CH3CHO --> CH4 + CO
In a particular experiment, the following kinetic data for the decomposition of acetaldehyde were obtained:
Time(s) 0 1000 2000 3000 4000
[CH3CHO] (M) 0.2480 0.1172 0.0767 0.0570 0.0454
What is the value of the rate constant at the temperature of this experiment?
Give an answer with concentration units of molarity and time units of seconds.

Answer 1

0.0495 M/s

Step-by-step explanation:

Rate of reaction (r) = Δ[
`CH_(3)CHO`]/Δt

Therefore, using the data given:

`r_(1)` = (0.2480-0.1172)M/(1000-0)s =
`1.308*10^(-4)` M/s

`r_(2)` = (0.1172-0.0767)M/(2000-1000)s =
`4.05*10^(-5)` M/s

Using the rate law:

`r = k*[CH_(3)CHO] ^(m)`

r is the rate of the reaction (M/s), k is the rate constant (M/s), and m is a number.

Therefore, we have:

`r_(1) = k*[0.1172]^(m)` (1)

`r_(2) = k*[0.0767]^(m)` (2)

Divide equation (1) by equation 2, we have:

`(r_(1) )/(r_(2) ) = [(0.1172)/(0.0767)] ^(m)`

using
`r_(1) = 1.308*10^(-4) M/s` and
`r_(2) = 4.05*10^(-5) M/s`

We have:

`(1.308*10^(-4) )/(4.05*10^(-5) ) = [(0.1172)/(0.0767) ]^(m)`

Thus:

3.2296 =
`1.528^(m)`

Taking log of both sides, we have

log (3.2296) = m*log (1.528), and m = 2.77 (approximately 3)

Therefore, using equation (1) to get the rate constant (k), we have:

`1.308*10^(-4)  (M/s) = k*[0.1172]^(2.77)`

Thus k = 0.0001308/0.00264 = 0.0495 M/s

Thus, the rate constant is 0.0495 M/s