Question

The rate constant for the reaction $C{O}_2+O{H}^{-}\to HC{O}_3^{-}$ in the aqueous solution is $4\times {10}^{-3}$ $Lmo{l}^{-1}se{c}^{-1}$. Calculate the number of mole of $C{O}_2$ and $O{H}^{-}$ used up per second when $\left[C{O}_2\right]$ and $\left[O{H}^{-}\right]$ are ${10}^{-6}$ and ${10}^{-1}$ mol L${}^{-1}$.

• A. $4\times {10}^{-10}mol{L}^{-1}{s}^{-1}$
• B. $4\times {10}^{-9}mol{L}^{-1}{s}^{-1}$
• C. $4\times {10}^{-5}mol{L}^{-1}{s}^{-1}$
• D. None of these

Answer 1

The correct option is A $4\times {10}^{-10}mol{L}^{-1}{s}^{-1}$

The unit of the rate constant indicates that it is a second-order reaction.
In the balanced chemical equation, the coefficients of $C{O}_2,O{H}^{-}$ and $HC{O}_3^{-}$ are 1 each.

Hence, their rates will be same and equal to the rate of the reaction.

$rate=k\left[C{O}_2\right]\left[O{H}^{-}\right]=4\times {10}^{-3}\times {10}^{-6}\times {10}^{-1}=4\times {10}^{-10}mol/\left(Ls\right)$