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For a first order gas phase reaction : ...

For a first order gas phase reaction :
`" "A(g)rarr2B(g)+C(g)`
`P_(0)` be initial pressure of A and `P_(t)` the total pressure at time 't' . Integrated rate equation is -

A

`(2.303)/(t)log((P_(0))/(P_(0)-P_(t)))`

B

`(2.303)/(t)log((2P_(0))/(3P_(0)-P_(t)))`

C

`(2.303)/(t)log((P_(0))/(2P_(0)-P_(t)))`

D

`(2.303)/(t)log((2P_(0))/(3P_(0)-P_(t)))`

Text Solution

Verified by Experts

The correct Answer is:
B
`" "A(g)rarr2B(g)+C(g)`
`{:("Initial pressure:",P_(0),0,0),("Pressure at time t:",P_(0)-p,2p,p):}`
Total pressure at time `t=P_(0)-p+2p+p=P_(t)`
or, `P_(t)=P_(0)+2p`
`thereforeP_(t)-P_(0)=2p " or, "p=(P_(t)-P_(0))/(2)`
`k=(2.303)/(t)log[(P_(0))/(P_(0)-p)]=(2.303)/(t)log[(P_(0))/(P_(0)-((P_(t)-P_(0))/(2)))]`
`=(2.303)/(t)log[(2P_(0))/(2P_(0)-P_(t)+P_(0))]=(2.303)/(t)log.((2P_(0))/(3P_(0)-P_(t)))`
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Knowledge Check

  • For the reaction A(g)rarrB (g) +C(g) , the rate constant is given as (P_(t) is initial pressure and P_(t) is pressure at time t )-

    A
    `k=(2.303)/(t)log.(P_(i))/(P_(t))`
    B
    `k=(2.303)/(t)log.(P_(i))/((2P_(i)-P_(t)))`
    C
    `k=(2.303)/(t)log.(2P_(i)-P_(t))/(P_(i))`
    D
    `k=(2.303)/(t)log.(2P_(i)-P_(t))/(2P_(i))`

  • Consider a first order gas phase decomposition reaction given below : A(g)rarrB(g)+C(g) The initial pressure of the system before decomposition of A was P_(i) After lapse of time 't' , total pressure of the system increased by x units and becomes 'P_(t)' . The rate constant k for the reaction is given as -

    A
    `k=(2.303)/(t)log.(P_(i))/(P_(i)-x)`
    B
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    C
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    D
    `k=(2.303)/(t)log.(P_(i))/(P_(i)+x)`

  • For the reaction AB(g)iffA(g)+B(g) , AB is 33% dissociated at a total pressure of P, then

    A
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    B
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    C
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    D
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