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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

For a given chemical reaction \(\gamma_{1} A +\gamma_{2} B \rightarrow \gamma_{3} C +\gamma_{4} D\) Concentration of \(C\) changes from \(10\, mmol\) appearance of \(D\) is \(1.5\) times the rate of disappearance of \(B\) which is twice the rate of disappearance \(A\). The rate of appearance of \(D\) has been experimentally determined to be \(9 \,m\,mol\) \(dm ^{-3} s ^{-1}\). Therefore the rate of reaction is \(......\,m\,mol\, dm ^{-3} \,s ^{-1}\). (Nearest Integer)

  1. A \(25\)
  2. B \(20\)
  3. C \(1\)
  4. D \(10\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(1\)

Step-by-step Solution

Detailed explanation

\(\gamma_{1} A +\gamma_{2} B \longrightarrow \gamma_{3} C +\gamma_{4} D\) Given : \(+\frac{ d [ D ]}{ dt }=\frac{-3}{2} \frac{ d [ B ]}{ dt }\) \(\Rightarrow \frac{-1}{2} \frac{ d [ B ]}{ dt }=\frac{+1}{3} \frac{ d [ D ]}{ dt }\)…