Question

For the reaction \( \frac{1}{2} A \rightarrow 2B \), rate of dissociation of \( A \) and rate of formation of \( B \) is related as follows:

• A. (i) \(-\frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt} \)
• B. (ii) \(-\frac{d[A]}{dt} = \frac{1}{4} \frac{d[B]}{dt} \)
• C. (iii) \(-\frac{d[A]}{dt} = \frac{d[B]}{dt} \)
• D. (iv) \(-\frac{d[A]}{dt} = 4 \frac{d[B]}{dt} \)

Hint:

In stoichiometry, the rates of reactants and products are inversely proportional to their respective coefficients in the balanced equation.

Answer 1

The Correct Option is A

For the reaction \( \frac{1}{2} A \rightarrow 2B \), the rate of reaction can be expressed in terms of the rate of disappearance of \( A \) and the rate of formation of \( B \): \[ -\frac{1}{2} \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}. \] Rearranging, we get: \[ -\frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}. \] This relation shows that the rate of dissociation of \( A \) is half the rate of formation of \( B \).