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For the reaction in equilibrium A harr B...

For the reaction in equilibrium `A harr B`
`([B])/([A])=4.0 xx 10^(8)" "(-d[A])/(dt)=2.3 xx 10^(6)S^(-1)[A]" "(-d)/(dt)[B]=K[B]` Thus, K is

A

`1.1 xx 10^(-15) S^(-1)`

B

`5.8 xx 10^(-3) S^(-1)`

C

`1.7 xx 10^(2) S^(-1)`

D

`9.2 xx 10^(14) S^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
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Knowledge Check

  • For the reaction N_(2)+3H_(2)to2NH_(3) , the rate (d[NH_(3)])/(dt)=2xx10^(-4)Ms^(-1) . Therfore the rate -(d[N_(2)])/(dt) is given

    A
    `10^(-4)M"sec"^(-1)`
    B
    `10^(4)M"sec^(-1)`
    C
    `10^(-2)M"sec"^(-1)`
    D
    `4xx10^(-4)M"sec"^(-1)`

  • For the reaction , N_(2)+3H_(2) to 2NH_(3), "if" (d[NH_(3)])/(dt)=2xx10^(-4)"mol"L^(-1)s^(-1), the value of (-d[H_(2)])/(dt) would be

    A
    `1xx10^(-4)"mol"L^(-1)s^(-1)`
    B
    `3xx10^(-4)"mol"L^(-1)s^(-1)`
    C
    `4xx10^(-4)"mol"L^(-1)s^(-1)`
    D
    `6xx10^(-4)"mol"L^(-1)s^(-1)`

  • The stepwise stability constants of a complex are given below. What is its overall reaction stability constant (beta_(4)) ? M + L hArr ML , K_(1) = 1.0 xx 10^(4) ML + L hArr ML_(2) , K_(2) = 1.0 xx 10^(3) ML_(2) + L hArr ML_(3) , K_(3) = 1.0 xx 10^(3) ML_(3) + L hArr ML_(4) , K_(4) = 1.0 xx 10^(2) (Overall reaction : M + 4L hArr ML_(4) )

    A
    `1.0 xx 10^(12)`
    B
    `12.1 xx 10^(3)`
    C
    `1.0 xx 10^(6)`
    D
    `1.0 xx 10^(8)`