Question

$H_2$$+$$I_2$$=$$2HI$. What will be the instantaneous rate of the reaction in terms of product?

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

Answer 1

The correct option is A $\frac{1}{2}$$\frac{d\left[HI\right]}{dt}$

Since the product concentration increases over time, there will be a positive sign in front of the rate expression. There are two moles of $HI$, so to get a unique value, or to bring all the molecules on the basis of same numerical value, we divide the instantaneous rate by the stoichiometric co-efficient. Hence the rate will be $\frac{1}{2}$$\frac{d\left[HI\right]}{dt}$