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

For the reaction in equilibrium A harr B ([B])/([A])=4.0 xx 10^(8)""(

1.	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
Answer» `1.1 xx 10^(-15) s^(-1)` `5.8 xx 10^(-3)s^(-1)` `1.7 xx 10^(2)s^(-1)` `9.2 xx 10^(14)s^(-1)` Solution :`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)`