The rate constant for the reaction:
`CO_(2) + overset(ɵ)(OH) rarr HCO_(3)^(ɵ)`
in the aqueous solution is `4 xx 10^(-3) L mol^(-1) s^(-1)`. Calculate the number of mole of `CO_(2)` and `overset(ɵ)(OH)` are `10^(-6)` and `10^(-1) mol L^(-1)`. Also predict the mole of `HCO_(3)^(ɵ)` formed per second.

Explanation

Rate of reaction `= (-d[CO_(2)])/(d t) = (-d[overset(ɵ)(OH)])/(dt) = (d[HCO_(3)])/(dt)`
Also the unit of `k` suggests it to be II order. Thus, rate of reaction
`= k[CO_(2)][overset(ɵ)(OH) ] = 4 xx 10^(-3) xx 10^(-6) xx 10^(-1)`
`= 4 xx 10^(-10) mol L^(-1) s^(-1)`