Question

Consider this reaction:

2Cl2O5 (g) → 2Cl2 (g) + 5O2 (g)

At a certain temperature it obeys this rate law.
rate = (6.48 M-1 • s-1)[Cl2O5]2
Suppose a vessel contains Cl2O5 at a concentration of 1.16 M. Calculate the concentration of Cl2O5 in the vessel 0.820 seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.

Answer 1

Answer : The concentration of $Cl_2O_5$ in the vessel 0.820 seconds later is, 0.16 M

Explanation :

The given reaction is:

$2Cl_2O_5(g)\rightarrow 2Cl_2(g)+5O_2(g)$

The rate law expression is:

$rate=(6.48M^{-1}s^{-1})[Cl_2O_5]^2$

The expression used for second order kinetics is:

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

where,

k = rate constant = $6.48M^{-1}s^{-1}$

t = time = 0.820 s

$[A_t]$ = final concentration = ?

$[A_o]$ = initial concentration = 1.16 M

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

$6.48\times 0.820=\frac{1}{[A_t]}-\frac{1}{1.16}$

$[A_t]=0.16M$

Therefore, the concentration of $Cl_2O_5$ in the vessel 0.820 seconds later is, 0.16 M