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NEET 2022
Exam
NEET
Year
2022
Subject
Chemistry
Chapter
Chemical Kinetics

NEET 2022 · Chemistry · Chemical Kinetics

For a chemical reaction 4 ~A+3 ~B arrow 6 C+9 D rate of formation...

For a chemical reaction 4 A+3 B→6C+9D4 \mathrm{~A}+3 \mathrm{~B} \rightarrow 6 \mathrm{C}+9 \mathrm{D} rate of formation of C\mathrm{C} is 6×10−2 mol L−1 s−16 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and rate of disappearance of A\mathrm{A} is 4×10−2 mol L−1 s−14 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}. The rate of reaction and amount of BB consumed in interval of 10 seconds, respectively will be:

Options

  1. A1×10−2 mol L−1 s−11 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 30×10−2 mol L−130 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
  2. B10×10−2 mol L−1 s−110 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 10×10−2 mol L−110 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
  3. C1×10−2 mol L−1 s−11 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 10×10−2 mol L−110 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
  4. D10×10−2 mol L−1 s−110 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 30×10−2 mol L−130 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}

Correct answer

A. 1×10−2 mol L−1 s−11 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 30×10−2 mol L−130 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}

Step-by-step solution

\begin{aligned}& \text{ Explanation: }4 \mathrm{~A}+3 \mathrm{~B} \longrightarrow6 \mathrm{C}+9 \mathrm{D}\\ & \qquad \begin{aligned}\text{ Rate of reaction }& =-\frac{1}{4} \frac{d[\mathrm{~A}]}{d t}=-\frac{1}{3} \frac{d[\mathrm{~B}]}{d t}\\ & =\frac{1}{6} \frac{d[\mathrm{C}]}{d t}=\frac{1}{9} \frac{d[\mathrm{D}]}{d t} \end{aligned} \end{aligned}Rateofreaction=RateofformationofRate of reaction = Rate of formation of\mathrm{C}..∴\quadr=16d[C]dt=16×6×10−2=1×10−2 mol L−1 s−1\begin{aligned} \therefore \quadr & =\frac{1}{6} \frac{d[\mathrm{C}]}{d t}=\frac{1}{6} \times6 \times10^{-2} \\ & =1 \times10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \end{aligned}Also,rateofreaction=RateofdisappearanceofAlso, rate of reaction = Rate of disappearance ofB r=−13d[ B]dt=d[ B]dt=3\timesr=3×10−2 mol L−1 s−1\begin{aligned}r & =-\frac{1}{3} \frac{d[\mathrm{~B}]}{d t}=\frac{d[\mathrm{~B}]}{d t}=3 \timesr \\ & =3 \times10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \end{aligned}ConsumptionofConsumption ofBin10secondsin 10 seconds=3 \times 10^{-2} \times 10=30=30\times10−210^{-2}\mathrm{~mol} \mathrm{~L}^{-1}$
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