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

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

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

$\frac {r_4}{r_2} =\frac {3}{2}$

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

$\frac {r_2}{r_1}= 2$
$r_4 = 1.5r_2 = 3r_1$
$\frac {d[C]}{dt}=$ 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 $1$.