Question

In a reaction $2HI \to {H_2} + {I_2},$the concentration of HI decreases from 0.5molL-1 to 0.4 mol L-1 in 10 minutes. What is the rate of reaction during this interval?
A. $5\times{10^{ - 3}}M{\min ^{ - 1}}$
B. $2.5\times{10^{ - 3}}M{\min ^{ - 1}}$
C. $5 \times{10^{ - 2}}M{\min ^{ - 1}}$
D. $2.5 \times{10^{ - 3}}M{\min ^{ - 1}}$

Answer 1

Hint: Rate of reaction is the change in concentration of reactants and products in unit time. Rate of reaction with respect to products is the increase in the concentration of products in unit time, Rate of reaction with respect to reactants is the decrease in the concentration of reactants in unit time.

Complete step by step answer:
$2HI \to {H_2} + {I_2},$
Rate of reaction with respect to reactants, d[A] is the change in concentration of reactants in unit time by its coefficient.
\[r = - \dfrac{1}{2}x\dfrac{{d[A]}}{{.dt}}\]
Change in concentration of reaction changes from 0.5 to 0.4 molL-1 ie, final concentration to initial concentration. Change in time is 10 min.
$ r = - \dfrac{1}{2}\times\dfrac{{d[0.4 - 0.5]mol{L^{ - 1}}}}{{10}} \\
r = - \dfrac{1}{2}\times\dfrac{{0.1mol{L^{ - 1}}}}{{10}} \\
r = 5\times{10^{ - 3}}M{\min ^{ - 1}} \\ $

Hence the option A is correct.

Note:
In a reaction, the reactants are converted into products that means the concentration of products increases and the concentration of reactants decreases, increase in concentration is indicated by +” and decrease in concentration is indicated by -“.