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=(1)/([A_t])-(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* 0.820=(1)/([A_t])-(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