The rate of formation for `C_(6)H_(6) + 3H_(2) underset(K_(b))overset(K_(f))hArr C_(6)H_(12)` for the forward reaction is first order with respect to `C_(6)H_(6)` and `H_(2)` each. Which are correct ?

Options

A. `K_(c )= (K_(f))/(K_(b))`
B. `K_(c) = ([C_(6)H_(12)])/([C_(6)H_(6)][H_(2)]^(3))`
C. `r_(f) = K_(f)[C_(6)H_(6)][H_(2)]`
D. `r_(b) = K_(b)[C_(6)H_(12)][H_(2)]^(-2)`

Answer

A, B, C, D

Explanation

`K_(c ) = (K_(r))/(K_(b)) = ([C_(6)H_(12)])/([C_(6)H_(6)][H_(2)]^(3)) = ([C_(6)H_(12)])/([C_(6)H_(6)][H_(2)][H_(2)]^(2))`
`r_(f) = K_(f)[C_(6)H_(6)][H_(2)], r_(b) = K_(b) xx` Unknown
at eq. `r_(f) = r_(b)K_(f)[C_(6)H_(6)][H_(2)] = K_(b) xx` Unknown `therefore " "` Unknown `= (K_(f))/(K_(b)) xx [C_(6)H_(6)][H_(2)]`
`r_(b) = K_(b) xx` Unknown `= K_(b) xx (K_(f))/(K_(b)) xx [C_(6)H_(6)][H_(2)]`
`= K_(f) xx [C_(6)H_(6)][H_(2)] = K_(b) xx [C_(6)H_(12)][H_(2)]^(-2) because K_(f) = (K_(b)[C_(6)H_(12)])/([C_(6)H_(6)][H_(2)]^(3))`