JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
The reaction between \(X\) and \(Y\) is first order with respect to \(X\) and zero order with respect to \(Y\).
| Experiment | \(\frac{[ X ]}{ mol \;L ^{-1}}\) | \(\frac{[ Y ]}{ mol\; L ^{-1}}\) | \(\frac{\text { Initial rate }}{ mol\; L ^{-1}\; min ^{-1}}\) |
| \(I\) | \(0.1\) | \(0.1\) | \(2 \times 10^{-3}\) |
| \(II\) | \(0.2\) | \(0.2\) | \(4 \times 10^{-3}\) |
| \(III\) | \(0.4\) | \(0.4\) | \(M \times 10^{-3}\) |
| \(IV\) | \(0.1\) | \(0.2\) | \(2 \times 10^{-3}\) |
- A \(20\)
- B \(30\)
- C \(40\)
- D \(50\)
Answer & Solution
Correct Answer
(C) \(40\)
Step-by-step Solution
Detailed explanation
\(r = k [ x ][ y ]^{0}= k [ x ]\) Using \(I\) and \(II\) \(\frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\left(\frac{ L }{0.1}\right) \Rightarrow L =0.2\) Using \(I\) and \(III\) \(\frac{M \times 10^{-3}}{2 \times 10^{-3}}=\frac{0.4}{0.1} \Rightarrow M=8\)…
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