For the set of reactions-i A-B k 1 - k 2 C- ii C-B D k 1 - A - - B - - k 2 - C - - k 3 - C - - B - is equal to-

The correct option is A d[ A ] dt K_ 1 [ A ] [ B ] K_ 2 [ C ] K_ 3 [ C ] [ B ] #160; = d[ A ] #160; dt #160; #160; = r ...

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For the set o...

Question

For the set of reactions,
(i) $A + B \begin{matrix} k_{1} ⇌ k_{2} \end{matrix} C$ , (ii) $C + B k_{3} - \to D$
$k_{1} [A] [B] - k_{2} [C] - k_{3} [C] [B]$ is equal to:

\frac{- d [A]}{d t}

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\frac{- d [B]}{d t}

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\frac{d [C]}{d t}

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\frac{d [D]}{d t}

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2 A \to B + 3 C; i f - \frac{d [A]}{d t} = k_{1} [A]^{2}; \frac{d [B]}{d t} = k_{2} [A]^{2}; \frac{d [C]}{d t} = k_{3} [A]^{2}

k_{1}, k_{2} a n d k_{3}

Which of the following expression is/are correct?

2 A ⟶ B + 2 C

- \frac{d [A]}{d t} = k_{1} [A]^{2}

- \frac{d [B]}{d t} = k_{2} [A]^{2}

- \frac{d [C]}{d t} = k_{3} [A]^{2}

k_{1}

k_{2}

k_{3}

2 A \to B + 3 C

- \frac{d [A]}{d t} = k_{1} [A]^{2}; \frac{d [B]}{d t} = k_{2} [A]^{2}; \frac{d [C]}{d t} = k_{3} [A]^{2}

k_{1}, k_{2}

k_{2}

2 A ⟶ B + 3 C

- \frac{d [A]}{d t} = k_{1} {[A]}^{2}; \frac{d [B]}{d t} = k_{2} {[B]}^{2}; \frac{d [A]}{d t} = k_{3} {[A]}^{2}

k_{1}, k_{2}

k_{3}

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