Question

For a reaction \(3A \rightarrow 2B\), the rate of reaction \(+\dfrac{d[B]}{dt}\) is equal to:

• A. \(-\dfrac{3}{2}\dfrac{d[A]}{dt}\)
• B. \(-\dfrac{2}{3}\dfrac{d[A]}{dt}\)
• C. \(-\dfrac{1}{3}\dfrac{d[A]}{dt}\)
• D. \(+\dfrac{2d[A]}{dt}\)

Hint:

For rate expression, always divide concentration change by stoichiometric coefficient.

Answer 1

The Correct Option is B

Concept:
For a balanced chemical reaction, rate of reaction is written by dividing the rate of change of concentration by the stoichiometric coefficient. For a general reaction: \[ aA \rightarrow bB \] the rate relation is: \[ -\frac{1}{a}\frac{d[A]}{dt} = +\frac{1}{b}\frac{d[B]}{dt} \] Reactant concentration decreases, so negative sign is used for reactant. Product concentration increases, so positive sign is used for product.

Step 1: Write the given reaction.
\[ 3A \rightarrow 2B \] Here, coefficient of \(A\) is \(3\), and coefficient of \(B\) is \(2\).

Step 2: Write the rate expression.
\[ -\frac{1}{3}\frac{d[A]}{dt} = +\frac{1}{2}\frac{d[B]}{dt} \]

Step 3: Find \(+\dfrac{d[B]}{dt}\).
Multiply both sides by \(2\): \[ +\frac{d[B]}{dt} = -\frac{2}{3}\frac{d[A]}{dt} \] Hence: \[ \boxed{+\frac{d[B]}{dt} = -\frac{2}{3}\frac{d[A]}{dt}} \] \[ \boxed{\text{(B)}} \]