Deducing Reaction Order from Pressure Data: This problem shows how the kinetics

Question

Deducing Reaction Order from Pressure Data: This problem shows how the kinetics of gas-phase reactions can be analyzed by monitoring the total pressure in a batch reactor. At 518 degree C, acetaldehyde vapor decomposes into methane and carbon monoxide according to CH_3CHO rightarrow CH_4 + CO. In a particular experiment carried out in a constant volume batch reactor, the initial pressure of acetaldehyde was 48 kPa, and the following increases of pressure (Delta P) were noted with increasing time: Verify using the integral method that the data are consistent with a second-order reaction. Calculate the value of the rate constant in pressure units.

Explanation / Answer

for second order reaction, -dCA/dt = KCA2

CA= PA/RT from gas law equation

-d(PA/RT)= K*(PA/RT)2

-dPA/dt = K* PA2/RT

R= 0.0821 L.atm/mole.K T= 518 deg.c= 518+273.15= 791.15K

1/RT= 1/(0.821* 791.15)= 0.0154

-dPA/dT= 0.0154*K PA2 = K' PA2,       K' is a constant =0.0154K

which on integration gives

1/PA= 1/PAO + K't

PA = pressure at any time and PAO= initial pressure ( both are in atm)

K' is rate constant

The plot of 1/PA Vs 1/t gives a straight line whose slope gives the value of rate constant if second order reaction has to be obeyed.

The plot along with the calculations are shown below

The slope is 0.005, the value of rate constant is 0.005/atm.sec

K= 0.0154/0.005 /M.sec) = 3.08 M.sec

The rate equation becomes

-dCA/dt= 3.08 CA2

PA Pa (Atm) t, sec 1/PA 48 0.47328 0 2.112914 43.5 0.42891 42 2.331491 38.1 0.375666 105 2.661939 30.1 0.296786 242 3.369431 22 0.21692 480 4.609994 15.2 0.149872 840 6.67236 10 0.0986 1440 10.14199

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