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

Answer:

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

Explanation:

 The second order equation is given by

$\frac{1}{R} = KT+\frac{1}{Ro}$

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

T= time = 6s

Ro= initial concentration = 0.044

$\frac{1}{R}$ = 0.8*6+ $\frac{1}{0.044}$

$\frac{1}{R}$ = 27.53

R= 1/27.53 = 0.0363