Question

A reaction is catalyzed by ${H^ + }$ ion. In presence of an acid $HA$ , the rate constant is $2 \times {10^{ - 3}}{\min ^{ - 1}}$ and in the presence of an acid $HB$ , the rate constant is $1 \times {10^{ - 3}}{\min ^{ - 1}}$ .Given $HA$ and $HB$ both being strong acid. We may conclude that.
The question has multiple correct answers
$A.$ Equilibrium constant is $2$
$B.$ $HA$ Is stronger acid than $HB$
$C.$ Relative strength of $HA$ and $HB$ is $2$
$D.$ $HA$ Is weaker acid than $HB$ and their relative strength is $0.5$

Answer 1

Rate constant is defined as the proportionality constant which explains the relationship between the molar concentration of the reactant and the rate of a chemical reaction. The rate constant is dependent upon temperature, catalyzed and activation energy. The rate equation of a chemical reaction when $X$ changes to $Y$ ,
$Rate = k\left[ X \right]$
Here, k is the rate constant.

Complete step by step answer:
In the Given question there are multiple correct answers. We have to find these correct answers. We know higher the value of the rate constant is higher the acidic strength. In given question we have given that rate constant in the presence of $HA$ is $2 \times {10^{ - 3}}{\min ^{ - 1}}$ and in presence of $HB$ the rate constant is $1 \times {10^{ - 3}}{\min ^{ - 1}}$ . In presence of $HA$ rate constant is higher. So we can say that $HA$ is strong acid than $HB$ .
Thus option $B$ is correct.
We can calculate the relative strength of two acids $HA$ and $HB$ by dividing the value of the rate constant of $HA$ by the rate constant of $HB$ .
Relative acidic strength$=$ Rate constant of $HA$ /Rate constant of $HB$
Relative acidic strength $= \dfrac{{2 \times {{10}^{ - 3}}}}{{1 \times {{10}^{ - 3}}}}$
$= 2$
Now, the relative strength is $2$ . Thus option $C$ is correct

The correct options are B\ & C

Additional information:
According to Arrhenius theory a strong acid is which release ${H^ + }$ ion more readily to the aqueous solution and strong base is which release $O{H^ - }$ ion more readily to the aqueous solution

Note:
The equilibrium constant in this equation cannot be calculated by using the rate of reaction of $HA$ and $HB$ because equilibrium constant is equal to the rate constant for the forward reaction divided by the rate constant for the reaction as no reaction is given which is in equilibrium . Hence we cannot calculate the equilibrium constant.