Answer:
E) rate of appearance of C = 0.45 M/s
rate of the reaction = 0.15 M/s
Explanation:
2A + B → 3C
Writing rate law for the reaction:
Rate of reaction = -[tex]\frac{1}{2}[/tex][tex]\frac{d[A]}{dt}[/tex] = -[tex]\frac{d[B]}{dt}[/tex] = [tex]\frac{1}{3}[/tex][tex]\frac{d[C]}{dt}[/tex] → equation 1
Given that the rate of disappearance of A is 0.3 M/s
⇒ - [tex]\frac{d[A]}{dt}[/tex] = 0.3 M/s
⇒Rate of reaction = - [tex]\frac{1}{2}[/tex][tex]\frac{d[A]}{dt}[/tex] = [tex]\frac{1}{2}[/tex]×0.3 M/s
⇒Rate of reaction = 0.15 M/s
From equation 1, [tex]\frac{d[C]}{dt}[/tex] = - [tex]\frac{3}{2}[/tex][tex]\frac{d[A]}{dt}[/tex] = [tex]\frac{3}{2}[/tex]×0.3 M/s
⇒[tex]\frac{d[C]}{dt}[/tex] = 0.45 M/s
or the rate of appearance of C = 0.45 M/s
