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Types of Equillibrium

Consider the ...

Question

Consider the elementary reaction sequence shown in the figure. Which of the following equations are correct?

\frac{d [A]}{d t} = - K_{1} [A] + K_{4} [D]

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\frac{d [C]}{d t} = K_{2} [B] + K_{3} [D]

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\frac{d [D]}{d t} = K_{4} [D] + K_{3} [D]

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\frac{d [B]}{d t} = K_{1} [A] - K_{2} [B]

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Solution

The correct options are
B $\frac{d [C]}{d t} = K_{2} [B] + K_{3} [D]$
D $\frac{d [B]}{d t} = K_{1} [A] - K_{2} [B]$
here,
$\frac{d [A]}{d t} = - K_{1} [A] - K_{4} [A]$
$\frac{d [C]}{d t} = K_{2} [B] + K_{3} [D]$
$\frac{d [D]}{d t} = K_{4} [A] - K_{3} [D]$
$\frac{d [B]}{d t} = K_{1} [A] - K_{2} [B]$

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Similar questions

Which of the following expression is/are correct?

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}

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}

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 N O (g) + 2 H_{2} (g) \to N_{2} (g) + 2 H_{2} O (g)

\frac{d [N_{2}]}{d t} = k_{1} [N O] [H_{2}]; \frac{d [H_{2} O]}{d t} = k_{2} [N O] [H_{2}]; \frac{- d [N O]}{d t} = k_{3} [N O] [H_{2}]; \frac{d [H_{2}]}{d t} = k_{4} [N O] [N_{2}]

k_{1}, k_{2}, k_{3}, k_{4}

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