Q48 / 180 · COMEDK 2025 (10 May Shift 1) · Chemistry
For a given reaction of the type $\dfrac{3}{5} X(a q) \rightarrow \dfrac{1}{2} Y(a q)+Z(g)$, the correct expression for the rate of disappearance of $X$ with reference to $Y$ is $\qquad$
- A$-d \dfrac{[X]}{d t}=+\dfrac{5}{6} d \dfrac{[Y]}{d t}$
- B$-d \dfrac{[X]}{d t}=+\dfrac{3}{10} d \dfrac{[Y]}{d t}$
- Correct answerC$-d \dfrac{[X]}{d t}=+\dfrac{6}{5} d \dfrac{[Y]}{d t}$
- D$-d \dfrac{[X]}{d t}=d \dfrac{[Y]^{1 / 2}}{d t}$
Solution
For a general chemical reaction $aA + bB \rightarrow cC + dD$, the rate of reaction is given by $-\dfrac{1}{a} \dfrac{d[A]}{dt} = -\dfrac{1}{b} \dfrac{d[B]}{dt} = \dfrac{1}{c} \dfrac{d[C]}{dt} = \dfrac{1}{d} \dfrac{d[D]}{dt}$.
Given the reaction $\dfrac{3}{5} X \rightarrow \dfrac{1}{2} Y + Z$, we apply the rate expression:
$-\dfrac{1}{3/5} \dfrac{d[X]}{dt} = \dfrac{1}{1/2} \dfrac{d[Y]}{dt}$
$-\dfrac{5}{3} \dfrac{d[X]}{dt} = 2 \dfrac{d[Y]}{dt}$
Rearranging to solve for the rate of disappearance of $X$, which is $-\dfrac{d[X]}{dt}$:
$-\dfrac{d[X]}{dt} = \dfrac{3}{5} \times 2 \dfrac{d[Y]}{dt}$
$-\dfrac{d[X]}{dt} = \dfrac{6}{5} \dfrac{d[Y]}{dt}$
Answer: $-d \dfrac{[X]}{d t}=+\dfrac{6}{5} d \dfrac{[Y]}{d t}$
Given the reaction $\dfrac{3}{5} X \rightarrow \dfrac{1}{2} Y + Z$, we apply the rate expression:
$-\dfrac{1}{3/5} \dfrac{d[X]}{dt} = \dfrac{1}{1/2} \dfrac{d[Y]}{dt}$
$-\dfrac{5}{3} \dfrac{d[X]}{dt} = 2 \dfrac{d[Y]}{dt}$
Rearranging to solve for the rate of disappearance of $X$, which is $-\dfrac{d[X]}{dt}$:
$-\dfrac{d[X]}{dt} = \dfrac{3}{5} \times 2 \dfrac{d[Y]}{dt}$
$-\dfrac{d[X]}{dt} = \dfrac{6}{5} \dfrac{d[Y]}{dt}$
Answer: $-d \dfrac{[X]}{d t}=+\dfrac{6}{5} d \dfrac{[Y]}{d t}$
Hence, the correct answer is option: 'C'.