Question

The rate constant for the decomposition of nitrogen dioxide NO2(g) LaTeX: \longrightarrow⟶ NO (g) + 1/2 O2(g) with a laser beam is 1.95 1/MLaTeX: \cdot⋅ min. Find the time, in seconds, needed to decrease 2.48 M of NO2 to 0.97 M. Hint: What is the order of the reaction? How can you determine that? Units of k?

Answer 1

Answer :

The time taken by the reaction is 19.2 seconds.

The order of reaction is, second order reaction.

Explanation :

The general formula to determine the unit of rate constant is:

$(Concentration)^{(1-n)}(Time)^{-1}$

Unit of rate constant                           Order of reaction

$Concentration/Time$                                        0

$Time^{-1}$                                                              1

$(Concentration)^{-1}Time^{-1}$                                 2

As the unit of rate constant is $M^{-1}min^{-1}$. So, the order of reaction is second order.

The expression used for second order kinetics is:

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

where,

k = rate constant = $1.95M^{-1}s^{-1}$

t = time = ?

$[A_t]$ = final concentration = 0.97 M

$[A_o]$ = initial concentration = 2.48 M

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

$1.95\times t=\frac{1}{0.97}-\frac{1}{2.48}$

$t=0.32min=0.32\times 60s=19.2s$

Therefore, the time taken by the reaction is 19.2 seconds.