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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
`A harr B, H_(c)=([B])/([A])=4.0 xx 10^(8)`
`(-d)/(dt)(A)` = Rate of forward
`= 2.3 xx 10^(6)S^(-1)[A]`
= rate constant of forward
Also `K_(c)=(K_(f))/(K_(b)) implies K_(b)=(K_(f))/(K_(c))=(2.3 xx 10^(6))/(4.0 xx 10^(8))`
`= 5.75 xx 10^(-2)s^(-1)`
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