Question

HI decomposes into its elements according to second-order kinetics. How long will it take for the concentration to decrease to 1.25 M from an initial concentration of 2.25 M? The rate constant, k, equals 1.6 × 10−3 M−1hr−1.
2 HI(g) → H2(g) + I2(g)

Answer 1

Answer : The time taken by the reaction is, $2.2\times 10^2hr$

Explanation :

The expression used for second order kinetics is:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

where,

k = rate constant = $1.6\times 10^{-3}M^{-1}s^{-1}$

t = time = ?

$[A_t]$ = final concentration = 1.25 M

$[A_o]$ = initial concentration = 2.25 M

Now put all the given values in the above expression, we get:

$(1.6\times 10^{-3})\times t=\frac{1}{1.25}-\frac{1}{2.25}$

$t=222.222hr\approx 2.2\times 10^2hr$

Therefore, the time taken by the reaction is, $2.2\times 10^2hr$