Question

The decomposition reaction 2 HI(g) → H2(g) + 12(g), is second order and has a rate constant equal to 1.6 x 10M's at 700 °C. If the initial concentration of HI in the container is 3.4 x 10-M, how many minutes will it take for the concentration to be reduced to 8.0 x 104 M?

Answer 1

Answer : The time taken by the reaction is 12714.46 minutes.

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 = $8.0\times 10^{-4}M$

$[A_o]$ = initial concentration = $3.4\times 10^{-2}M$

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

$(1.6\times 10^{-3})\times t=\frac{1}{8.0\times 10^{-4}}-\frac{1}{3.4\times 10^{-2}}$

$t=762867.6471s=12714.46min$      (1 min = 60 s)

Therefore, the time taken by the reaction is 12714.46 minutes.