The rate of a reaction is expressed in different ways as follows-1-2 d-C-dt -1-3 d-D-dt -1-4 d-A-dt - d-B-dtThe reaction is

B+3D→ 4A+2C d[B]/dt= 1/3d[D]/dt=1/4d[A]/dt=1/2d[C]/dt ...

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Byju's Answer

Standard XII

Chemistry

Rate of Reaction

The rate of a...

Question

The rate of a reaction is expressed in different ways as follows:
$\frac{1}{2} (\frac{d [C]}{d t}) = - \frac{1}{3} (\frac{d [D]}{d t}) = \frac{1}{4} (\frac{d [A]}{d t}) = - (\frac{d [B]}{d t})$
The reaction is

4 A + B \to 2 C + 3 D

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B + 3 D \to 4 A + 2 C

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4 A + 2 B \to 2 C + 3 D

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B + (1 / 2) D \to 4 A + 3 C

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Solution

The correct option is B $B + 3 D \to 4 A + 2 C$
$B + 3 D \to 4 A + 2 C \frac{- d [B]}{d t} = \frac{- 1}{3} \frac{d [D]}{d t} = \frac{1}{4} \frac{d [A]}{d t} = \frac{\frac{1}{2} d [C]}{d t}$

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Q. The rate of a reaction is expressed in different ways as follows:

\frac{1}{2} (\frac{d [C]}{d t}) = - \frac{1}{3} (\frac{d [D]}{d t}) = \frac{1}{4} (\frac{d [A]}{d t}) = - (\frac{d [B]}{d t})

The reaction is

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+ \frac{1}{2} \frac{d [C]}{d t} = - \frac{1}{3} \frac{d [D]}{d t} = + \frac{1}{4} \frac{d [A]}{d t} = - \frac{d [B]}{d t}

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= \frac{1}{2} \frac{d [C]}{d t} = - \frac{1}{3} \frac{d [D]}{d t} = \frac{1}{4} \frac{d [A]}{d t} = - \frac{d [B]}{d t}

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Consider the chemical reaction

$N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} (g) .$

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The rate of a reaction is expressed in different ways as follows: 12(d[C]dt)=−13(d[D]dt)=14(d[A]dt)=−(d[B]dt) The reaction is

The correct option is B B+3D→4A+2CB+3D→4A+2C−d[B]dt=−13d[D]dt=14d[A]dt=12d[C]dt

The rate of a reaction is expressed in different ways as follows:
$\frac{1}{2} (\frac{d [C]}{d t}) = - \frac{1}{3} (\frac{d [D]}{d t}) = \frac{1}{4} (\frac{d [A]}{d t}) = - (\frac{d [B]}{d t})$
The reaction is

The correct option is B $B + 3 D \to 4 A + 2 C$
$B + 3 D \to 4 A + 2 C \frac{- d [B]}{d t} = \frac{- 1}{3} \frac{d [D]}{d t} = \frac{1}{4} \frac{d [A]}{d t} = \frac{\frac{1}{2} d [C]}{d t}$