IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry - Chemical Kinetics-Help needed

harsha_27

Some PH3 (g) is introduced into a flask at 600C containing an inert gas.PH3 proceeds to decompose into P4(g) and the reaction goes to completion. The total pressure is given below as a function of time.Find the order of the reaction and calculate the rate constant.

Time(sec.) 0 60 120 infinity

Press.(mm) 262.40 272.90 275.51 276.40

The answer given is First order and K = 0.023 /sec.

Please solve this.My answer for K isn't matching.

Don't worry about rates,I'll give for each reply....ok

karthik2007

Can you give the complete reaction?

is it 4PH3 ------> P4 + 6H2

harsha_27

Yeah.

But the question framer has not mentioned the reaction.I've typed the complete question.No data missing.Typed it as it is.

kartick

4PH3 + INERT GAS- P4(g) + 6H2(g)+ INERT GAS

T=0 P P` 0 0

T=t (P-A) P` A\4 6A\4 P`

T=infinity 0 P` P\4 6P\4 P`

T=0 P=262.40=P(PH3)+P(INERT GAS)

T=60 P=272.90=P(PH3 LEFT)+P(P4)+P(H2)+P(INERTGAS)

T=120 P=275.51=P(PH3 LEFT)+P(P4)+P(H2)+P(INERTGAS)

T=infinity P=276.40=P(P4)+P(H2)+P(INERT GAS)

P\4+6P\4+P`=276.40

7P+4P`=276.40 * 4

ALSO AT T=0 P+P`=262.40

P=18.66 MM P`=273.74 MM

AT T=60 PRESSURE=272.90=P-A + P` + A/4 + 6A\4

272.90=18.66-A+243.74+7A\4

A=14mm K=2.303\T LOG ( P(PH3)AT T=0 DIVIDED BY P(PH3)AT TIME T=60)

=2.303\60 LOG18.66\18.66-14=2.31\100 PER SEC

AT T=120 PRESSURE=271.51=P-A`+P`+A`\4+6A`\4

271.51=18.66-A`+243.74+7A`\4

A`=17.48MM

K=2.303\120 LOG(18.66 DIVIDED BY 18.66-17.48) =2.30/100 PER SECOND

harsha_27

Thanks kk.......Thanks for typing every step.......instead u could have called me w/o typing everything here.......Anyway thanks for ur efforts....

karthik2007

Excellent work man... The inert gas idea didnt strike me at all

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