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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

Respuesta :

Answer:

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

Explanation:

 The second order equation is given by

[tex]\frac{1}{R} = KT+\frac{1}{Ro}[/tex]

k= rate constant= 0.8[tex]M^-1s^-1[/tex]

T= time = 6s

Ro= initial concentration = 0.044

[tex]\frac{1}{R}[/tex] = 0.8*6+ [tex]\frac{1}{0.044}[/tex]

[tex]\frac{1}{R}[/tex] = 27.53

R= 1/27.53 = 0.0363

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