Consider the following reaction, $4 A + B ⟶ 2 C + 2 D$ The true statements about this reaction are

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the rates of formation of C and D are equal

the rate of appearance of C is one-half of the rate of disappearance of B

the rate of disappearance of B is one-fourth of the rate of disappearance of A

the rate of formation of D is one -half of the rate of consumption of A

answer is B.

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Detailed Solution

For the reaction, $4 A + B ⟶ 2 C + 2 D$ ,
Rate $= - \frac{1}{4} \frac{d [A]}{dt} = - \frac{d [B]}{dt} = \frac{1}{2} \frac{d [C]}{dt} = \frac{1}{2} \frac{d [D]}{dt}$
So, rate of appearance of C is twice of rate of disappearance of B. All other statements are true.