Question

At 400 K oxalic acid decomposes according to the reaction:
H2C2O4(g)→CO2(g)+HCOOH(g)
In three separate experiments, the intial pressure of oxalic acid and final total pressure after 20,000 s is measured.

Experiment	1	2	3
P H2C2O4 at t = 0	65.8	92.1	111
PTotal at t = 20,000 s	96.4	135	163

Find the rate law of the reaction.

At 400  oxalic acid decomposes according to the reaction:

In three separate experiments, the intial pressure of oxalic acid and final total pressure after 20,000  is measured.

Experiment	1	2	3
at  = 0	65.8	92.1	111
at  = 20,000	96.4	135	163

Find the rate law of the reaction.

Rate= 2.33×10⁻⁵ s−1PH2C2O4

Rate= 3.82×10⁻⁵ s−1PH2C2O4

Rate= 9.55×10⁻⁶ s−1PH2C2O4

Rate= 5.73×10⁻⁵ s−1PH2C2O4

Answer 1

given data points suggests that Ptotal/Pat t=0 = 96.8/65.4= 1.465 from 1st data point

from 2nd data point, Ptotal/P at t=0 = 135/92.1= 1.465, for the 3rd data point, Ptotal/P at t=0 is = 163/111=1.465

so the all the data points sasity the equal ratio and hence the reaction is 1st order for 1st order reaction given

H2C2O4(g)→CO2(g)+HCOOH(g)

let Po= initial pressure and change= x

hence pressrue at any time , H2C2O4= Po-x, CO2= x and HCOOH=x

hence total presure = Po-x+x+x= Po+x= 1.465Po

x=0.465Po , Po-x =0.535Po

hecne the integrated expression for -dP/dt= K becomes

ln(P/Po)= -Kt

P=0.465P0, Po=Po=initial presure , t= 20000sec

K=- ln(0.465)/20000= 3.82*10^-5/sec

hence the rate expression beomes rate= 3.82*10^-5PH2C2O4. ( B is correct)