Question

$2KCl{O}_3\to 2KCl$$+$$3O_2$. What will be the instantaneous rate of the reaction in terms of the reactant?

• A. $\frac{1}{2}$$\frac{d\left[KCl{O}_3\right]}{dt}$
• B. $-$$\frac{1}{2}$$\frac{d\left[KCl{O}_3\right]}{dt}$
• C. $\frac{d\left[KCl{O}_3\right]}{dt}$
• D. $-$$\frac{d\left[KCl{O}_3\right]}{dt}$

Answer 1

The correct option is B $-$$\frac{1}{2}$$\frac{d\left[KCl{O}_3\right]}{dt}$

Since $KCl{O}_3$ is the reactant, the concentration will decrease with time. Hence, there will be a negative sign in front of the rate expression. There are $2$ moles of $KCl{O}_3$. To bring all the molecules on the same numerical basis or to get a unique value, we divide the rate by the stoichiometric co-efficient.

Hence, option B is correct.