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Given that for a reaction of nth order, ...

Given that for a reaction of nth order, the integrated rate equation is:
`K=1/(t(n-1))[1/C^(n-1)-1/C_(0)^(n-1)]`, where `C` and `C_(0)` are the concentration of reactant at time `t` and initially respectively. The `t_(3//4)` and `t_(1//2)` are related as `t_(3//4)` is time required for C to become `C_(1//4`) :

A

`t_(3//4)=t_(1//2) [2^((n-1))-1]`

B

`t_(3//4)=t_(1//2) [2^((n-1))+1]`

C

`t_(3//4)=t_(1//2) [2^((n+1))+1]`

D

`t_(3//4)=t_(1//2) [2^((n-1))+1]`

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The correct Answer is:
D
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Knowledge Check

  • Given that for a reaction of order n. the intergrated form of the rate equation is k= (1)/(t(n-1))[(1)/(C^(n-1))-(1)/(C_(0)^(n-1))] where C_(0) and C are the values after time t. What is the relationship between t_(3//4) and t_(1//2) where t_(3//4) is the time required for C to become 1//4C_(0)-

    A
    `t_(3//4) = t_(1//2)[2^(n-1) +1]`
    B
    `t_(3//4) =t_(1//2)[2^(n-1)-1]`
    C
    `t_(3//4) =t_(1//2) [2^(n+1) -1]`
    D
    `t_(3//4) =t_(1//2) [2^(n+1)+1]`

  • Given that for a reaction of order n, the integrated from of the rate equation is : k=(1)/(t(n-))[10/(C^(n-1))-(1)/(C_(0)^(n-1))] where C_(0) and C are the values of the reactant concentration at the start and after tme t. What is the relationship between t_(3//4) and t_(1//2) where t_(3//4) is the time required for C to become (1)/(4)C_(0) ?

    A
    `t_(3//4)=t_(1//2)[2^(n-1)+1]`
    B
    `t_(3//4)=t_(1//2)[2^(n-1)-1]`
    C
    `t_(3//4)=t_(1//2)[2^(n+1)-1]`
    D
    `t_(3//4)=t_(1//2)[2^(n+1)+1]`

  • Given that for a reaction of order n the integrated form of the rate equation is k=1/(t(n-1))[1/C_(n-1) -1/(C_0^(n-1))] where C_0 and C are the values of the reactant concentration at the start and after time t . What is the relationship between t_(3/4) and t_(½) ?

    A
    `t_(3//4)=t_(1///2) [2^(n-1)+1]`
    B
    `t_(3//4)=t_(1///2) [2^(n-1)-1]`
    C
    `t_(3//4)=t_(1///2) [2^(n-1)-1]`
    D
    `t_(3//4)=t_(1///2) [2^(n+1)+1]`

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