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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 \(\frac{[ X ]}{ mol \;L ^{-1}}\) \(\frac{[ Y ]}{ mol\; L ^{-1}}\) \(\frac{\text { Initial rate }}{ mol\; L ^{-1}\; min ^{-1}}\)
\(I\) \(0.1\) \(0.1\) \(2 \times 10^{-3}\)
\(II\) \(0.2\) \(0.2\) \(4 \times 10^{-3}\)
\(III\) \(0.4\) \(0.4\) \(M \times 10^{-3}\)
\(IV\) \(0.1\) \(0.2\) \(2 \times 10^{-3}\)
Examine the data of table and calculate ratio of numerical value of \(M\)

  1. A \(20\)
  2. B \(30\)
  3. C \(40\)
  4. D \(50\)
Verified Solution

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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