Question

The decomposition of nitrous oxide at 565o

c.2n2o → 2n2 + o2 is second order in n2o with a rate constant of 1.10 x 10‐3 m‐1 s‐1 . if the reaction is initiated with [n2o] equal to 0.108 m, what will its concentration be after 1250 s have elapsed?

Answer 1

Answer: -

Concentration of nitrous oxide after 1250 s have elapsed A = 0.723 M

Explanation: -

The second order rate equation is

$\frac{1}{[A]}$ = $\frac{1}{[A0]}$ + kt

Ao = Initial amount of nitrous oxide taken = 0.108 M

k = rate constant = 1.10 x 10⁻³ M‐1 s‐1

t = time passed = 1250 s

Plugging in the values into the formula we get

$\frac{1}{[A]}$ = $\frac{1}{[108]}$ + 1.10 x 10⁻³ x 1250

                                    =  $\frac{1}{[108]}$ + 1.375

                                   = 1.384

Concentration of nitrous oxide  after 1250 s have elapsed A = 0.723 M