Question

For a given chemical reaction
\(γ_1A + γ_2B → γ_3C + γ_4D\)
Concentration of C changes from 10 mmol dm^–3 to 20 mmol dm^–3 in 10 seconds. Rate of 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 mmol dm^–3 s^–1. Therefore, the rate of reaction is _____ mmol dm^–3 s^–1. (Nearest Integer)

Answer 1

Correct Answer: 1

\(\frac {1}{r_1}(\frac {-d[A]}{dt}) = \frac {1}{r_2}(\frac {-d[B]}{dt}) = \frac {1}{r_3}(\frac {-d[C]}{dt}) =\frac {1}{r_4}(\frac {d[D]}{dt)}\)

\(\frac {d[D]}{dt} = \frac {r_4}{r_2}(\frac {-d[B]}{dt})\)

\(\frac {r_4}{r_2} =\frac {3}{2}\)

\(\frac {d[B]}{dt} = \frac {r_2}{r_1}(\frac {-d[B]}{dt})\)

\(\frac {r_2}{r_1}= 2\)
\(r_4 = 1.5r_2 = 3r_1\)
\(\frac {d[C]}{dt}=\) 1 m.mol dm^–3 sec^–1

\(\frac {d[D]}{dt}=\) 9 m.mol dm^–3 sec^–1

\(\frac {d[D]}{dt} = \frac {r_4}{r_3}.\frac {d[C]}{t}\)

\(\frac {r_4}{r_3} = 9\)
\(r_4 = 9r_3 = 3r_1\)
\(r_1 = 3r_3\)
\(3r_3A + 6r_3B → r_3C + 9r_3D\)
So, rate of reaction = \(\frac 19 \times 9\) m.mol dm^–3 sec^–1
= \(1\) m.mol dm^–3 sec^–1

So, the answer is \(1\).