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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

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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

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