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Physical Chemistry

A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is convertedinto CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, thevalue of K is

nikhil

12 Years agoGrade

4 Answers

Sunil Kumar FP

11 Years ago

CO2 +C ------2CO
initial presasure=.5atm
final pressure on addition of graphite
CO2=.5-X
CO=2X
total pressure at equilibrium=.5+x
but the total pressure at equilibrium= .8atm
,5+x=.8atm
x=.3atm
the value of reaction constant,K=[CO]^2/[CO2]=.6^2/.3=1.2

Elsa

10 Years ago

:)/(0.5-0.3)] =1.8atm ²]=[0.6₂/[CO²=0.5-XCO=2X Total pressure at equilibrium= 0.5+xbut the total pressure at equilibrium= 0.8atm0.5+x=0.8atm; x=0.3atm.The value of reaction constant, K=[CO]₂2CO Initial presasure=0.5atm Final pressure on addition of graphite: CO ------+C₂CO

payal kendre

8 Years ago

CO2 pressure decreases let it be `x`then increase in pressure due to CO after reaction is 2xtotal pressure = (0.5-x) + 2x = 0.8 ATMX=0.3 ATMthus pressure of CO2 = 0.2and CO= 0.6K = P^2 CO ÷ P CO2 = 1.8

Kushagra Madhukar

5 Years ago

Dear student,

Please find the solution to your problem below.

CO₂ + C → 2CO

initial presasure = 0.5 atm

final pressure on addition of graphite

CO₂ = 0.5 – x

CO = 2x

total pressure at equilibrium = 0.5 + x

but the total pressure at equilibrium = 0.8 atm

0.5 + x = 0.8 atm

x = 0.3 atm

Hence, at equilibrium pressure of CO₂ = 0.2 atm and CO = 0.6 atm

the value of reaction constant,K=[CO]²/[CO₂] = 0.6²/0.3 = 1.2

Thanks and regards,

Kushagra

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