Question

19. The reaction 2NOBr (g) → 2 NO (g) + Br2 (g) is a second-order reaction with a rate constant of 0.80 M-1s-1 at 11 °C. If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 6.0 seconds is ________. A) 0.0276 M B) 0.0324 M C) 0.0363 M D) 0.0348 M E) 0.0402 M

Answer 1

(C)The concentration of NOBr after 6 seconds is 0.0363

Step-by-step explanation:

The second order equation is given by

`(1)/(R) = KT+(1)/(Ro)`

k= rate constant= 0.8
`M^-1s^-1`

T= time = 6s

Ro= initial concentration = 0.044

`(1)/(R)` = 0.8*6+
`(1)/(0.044)`

`(1)/(R)` = 27.53

R= 1/27.53 = 0.0363