Question 11 Mark
For the reaction $A + 3B \rightarrow 2C + 2D,$ the concentration of $A$ changes from $0.150 M$ to $0.0135 M$ in $1$ min. The rate of formation of $C$ in $mol/L/s$ is:
Answer
View full question & answer→$(c)\ 5 \times 10^{-5}$
Explanation: rate $=-\frac{d[A]}{d t}=-\frac{1}{3} \frac{d[B]}{d t}=+\frac{1}{2} \frac{d[C]}{d t}=+\frac{1}{2} \frac{d[D]}{d t}$
$\Rightarrow \frac{d[A]}{d t}=-\frac{1}{2} \frac{d[C]}{d t}$
$\Rightarrow \frac{d[C]}{d t}=2 \frac{d[A]}{d t}$
$=-\frac{2 \times(0.0135-0.0150)}{1 \times 60}$
$=5 \times 10^{-5} mol\ L^{-1}\ s^{-1}$
rate $=5 \times 10^{-5} mol\ L^{-1}\ s^{-1}$
Explanation: rate $=-\frac{d[A]}{d t}=-\frac{1}{3} \frac{d[B]}{d t}=+\frac{1}{2} \frac{d[C]}{d t}=+\frac{1}{2} \frac{d[D]}{d t}$
$\Rightarrow \frac{d[A]}{d t}=-\frac{1}{2} \frac{d[C]}{d t}$
$\Rightarrow \frac{d[C]}{d t}=2 \frac{d[A]}{d t}$
$=-\frac{2 \times(0.0135-0.0150)}{1 \times 60}$
$=5 \times 10^{-5} mol\ L^{-1}\ s^{-1}$
rate $=5 \times 10^{-5} mol\ L^{-1}\ s^{-1}$
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is 2-Methylbutan-2-ol
Prop-1-en-2-ol (A) acetone are tautomers.