Question

The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when [NOBr]0 5 0.072 M and [NOBr]0 5 0.054 M.

Answer 1

(a)

`0.0342M`

(b)

`t_(1/2)=17.36s\\t_(1/2)=23.15s`

Step-by-step explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

`(1)/([NOBr])=kt +(1)/([NOBr]_0)\\(1)/([NOBr])=(0.8)/(M*s)*22s+(1)/(0.086M)=(29.3)/(M)\\`

`[NOBr]=(1)/(29.2/M)=0.0342M`

(b) Now, for a second-order reaction, the half-life is computed as shown below:

`t_(1/2)=(1)/(k[NOBr]_0)`

Therefore, for the given initial concentrations one obtains:

`t_(1/2)=(1)/((0.80)/(M*s)*0.072M)=17.36s\\t_(1/2)=(1)/((0.80)/(M*s)*0.054M)=23.15s`

Best regards.

Answer 2

Step-by-step explanation:

2NOBr(g) --> 2NO(g) 1 Br2(g)

Rate constant, k = 0.80

a) Initial concentration, Ao = 0.086 M

Final Concentration, A = ?

time = 22s

These parameters are connected with the equation given below;

1 / [A] = kt + 1 / [A]o

1 / [A] = 1 / 0.086 + (0.8 * 22)

1 / [A] = 11.628 + 17.6

1 / [A] = 29.228

[A] = 0.0342M

b) t1/2 = 1 / ([A]o * k)

when [NOBr]0 5 0.072 M

t1/2 = 1 / (0.072 * 0.80)

t1/2 = 1 / 0.0576 = 17.36 s

when [NOBr]0 5 0.054 M

t1/2 = 1 / (0.054 * 0.80)

t1/2 = 1 / 0.0432 = 23.15 s