Q39 / 180 · COMEDK 2025 (25 May Shift) · Chemistry
In the following reaction $x A \rightarrow y B$,
$\log _{10}\left[-\dfrac{d[A]}{d t}\right]=\log _{10}\left[\dfrac{d[B]}{d t}\right]+0.3010$
The relation between x and y is: [Given: $\log 2=0.3010, \log 4=0.6020$ ]
- Correct answerA$x=2 y$
- B$x= y$
- C$x=3 y$
- D$x=\dfrac{y}{2}$
Solution
For the reaction $x A \rightarrow y B$, the rate of reaction is given by:
$-\dfrac{1}{x} \dfrac{d[A]}{dt} = \dfrac{1}{y} \dfrac{d[B]}{dt}$
This implies $-\dfrac{d[A]}{dt} = \dfrac{x}{y} \dfrac{d[B]}{dt}$
Taking $\log_{10}$ on both sides:
$\log_{10} \left( -\dfrac{d[A]}{dt} \right) = \log_{10} \left( \dfrac{x}{y} \dfrac{d[B]}{dt} \right)$
$\log_{10} \left( -\dfrac{d[A]}{dt} \right) = \log_{10} \left( \dfrac{d[B]}{dt} \right) + \log_{10} \left( \dfrac{x}{y} \right)$
Comparing this with the given equation $\log_{10} \left( -\dfrac{d[A]}{dt} \right) = \log_{10} \left( \dfrac{d[B]}{dt} \right) + 0.3010$, we get:
$\log_{10} \left( \dfrac{x}{y} \right) = 0.3010$
Since $0.3010 = \log_{10} 2$, we have:
$\log_{10} \left( \dfrac{x}{y} \right) = \log_{10} 2$
$\dfrac{x}{y} = 2 \Rightarrow x = 2y$
Answer: $x=2 y$
$-\dfrac{1}{x} \dfrac{d[A]}{dt} = \dfrac{1}{y} \dfrac{d[B]}{dt}$
This implies $-\dfrac{d[A]}{dt} = \dfrac{x}{y} \dfrac{d[B]}{dt}$
Taking $\log_{10}$ on both sides:
$\log_{10} \left( -\dfrac{d[A]}{dt} \right) = \log_{10} \left( \dfrac{x}{y} \dfrac{d[B]}{dt} \right)$
$\log_{10} \left( -\dfrac{d[A]}{dt} \right) = \log_{10} \left( \dfrac{d[B]}{dt} \right) + \log_{10} \left( \dfrac{x}{y} \right)$
Comparing this with the given equation $\log_{10} \left( -\dfrac{d[A]}{dt} \right) = \log_{10} \left( \dfrac{d[B]}{dt} \right) + 0.3010$, we get:
$\log_{10} \left( \dfrac{x}{y} \right) = 0.3010$
Since $0.3010 = \log_{10} 2$, we have:
$\log_{10} \left( \dfrac{x}{y} \right) = \log_{10} 2$
$\dfrac{x}{y} = 2 \Rightarrow x = 2y$
Answer: $x=2 y$
Hence, the correct answer is option: 'A'.