Sponsor Area

Chemical Kinetics

Question
CBSEENCH12006250

The gas phase decomposition of acetaldehyde.
CH3CHO(g)   CH4(g) + CO(g)

at 680 K is observed to follow the rate expression.

Rate-dCH3CHOdt = kCH3CHO3/2

If the rate of decomposition is followed by monitoring the partial pressure of the acetaldehyde we can express the rate as

-dPCH3CHOdt = kPCH3 CHO3/2

If the pressure is measured in atmosphere and time in minutes, then
(a) What are the units of the rate of reaction?
(b) What are the units of rate constant k?

Solution
Rate=-dPCH3CHOdt = kPCH3 CHO3/2

(i) Unit of rate of reaction

                 = AtmosphereUnit of time = atm min-1                              (If time is in minutes)


(ii) 
            k = Rate(PCH3CHO)3/2 = atm min-1(atm)3/2    = atm-1/2 min-1.