Question

The second-order decomposition of NO2 has a rate constant of 0.255 M-15-1. How much NO2 decomposes in 4.00 s if the initial concentration of NO2 (1.00 L volume) is 1.33 M? A) 1.8 mol B) 0.85 mol C) 0.48 mol D) 0.77 mol

Answer 1

Answer:

Option D, Concentration of NO2 decomposes after 4.00 s = 0.77 mol

Explanation:

$rate\;constant = 0.255\;M^{-1}s{-1}$

Time (t) = 4.00\;s

Initial concentration of NO2 = 1.33 M

Integrated law for second order reaction:

$\frac{1}{[A]}=\frac{1}{[A]_0} =kt$

Where, [A] = Concentration after time, t

[A]0 = Intitial concentration, k = rate constant, t = time

On substituting values in the above

$\frac{1}{[A]}=\frac{1}{1.33} =0.255 \times 4.00$

$\frac{1}{[A]} =1.772$

[A] = 0.5644 M

Concentration of NO2 decomposes after 4.00 s = 1.33 - 0.5644 = 0.7656 M

No. of mole = Molarity * volume

                    = 0.7656 * 1

                    = 0.7656 mol 0r 0.77 mol