The rate law for the reaction 2 NOBrg 2 NOg Br2g at some te

The rate law for the reaction 2 NOBr(g) 2 NO(g) + Br2(g) at some temperature is the following. RatedNOBr] dt (a) If the half-life for this reaction is 2.10 s when [NOBrlo 0.935 M, calculate the value of k for this reaction. L mol-1 s-1 (b) How much time is required for the concentration of NOBr to decrease to 0.362 M?

Solution

a)

Answer

k = 0.5093L mol^-1s^-1

Explanation

2NOBr(g)- - - - - - > 2NO(g) + Br2(g)

This second order reaction

rate = k[NOBr]

For second order reaction

Half life, t_1/2 = 1/k[A]₀

k = 1/t_1/2[A]₀

   _{k = 1/(2.10s × 0.935M)}

k = 0.5093 L mol^-1s^-1

b)

Answer

3.32s

Explanation

Integrated form of second order rate equation is

1/[A]_t = kt + 1/[A]₀

   1/0.362M= (0.5093M^-1s^-1× t) + 1/0.935M

2.7624M^-1= (0.5093M^-1s^-1 × t) + 1.0695M^-1

t× 0.5093M-1s^-1 = 1.6929M^-1

t = 1.6929M^-1/0.5093M^-1s^-1

   t= 3.32s