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At 518^(@)C the rate of decomposition o...

At `518^(@)C` the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 torr, was `1.00"torr".^(-1)` when 5% had reacted and `0.5"torr.s"^(-1)` when 33% had reacted. The order of the reaction is -

A

1

B

0

C

2

D

3

Text Solution

Verified by Experts

The correct Answer is:
C
Rate = `k[P_(CH_(3)CHO)]^(n)`
After 5% completion of reaction,
`P_(CH_(3)CHO)=363-363xx(5)/(100)=363xx0.95` torr
At this stage, `1"torr".s^(-1)=k(363xx0.95" torr")^(n)" " ...[1]`
After 33% completion of reaction
`P_(CH_(3)CHO)=363-363xx(33)/(100)=363xx0.67` torr
`therefore 0.5 " torr".s^(-1)=k(363xx0.67" torr")^(n)" "...[2]`
From equations (1) and (2) we get- `2=((0.95)/(0.67))^(n)=(1.4179)^(n)`
or,` log2=log (1.433)^(n)`
or, `n~~2`
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