The rate of reaction is expressed as rate - 1 2 d-C- dt - 1 - 3 d-D- - dt - 1 - 4 d-A- - dt - d-B- - dt- The reaction is

Given Rate expression is #160; 1/2d[c]/dt = 1/3d[D]/dt = 1/4d[A]/dt = d[B]/dt Reaction is : B+3D → 4A+2C ...

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Standard X

Chemistry

Factors Affecting the Rate of a Chemical Reaction

The rate of r...

Question

The rate of reaction is expressed as rate
$= \frac{1}{2} \frac{d [C]}{d t} = - \frac{1}{3} \frac{d [D]}{d t} = \frac{1}{4} \frac{d [A]}{d t} = - \frac{d [B]}{d t}$
. The reaction is

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Solution

given Rate expression is
$\frac{1}{2} \frac{d [c]}{d t} = \frac{- 1}{3} \frac{d [D]}{d t} = \frac{1}{4} \frac{d [A]}{d t} = \frac{- d [B]}{d t}$
Reaction is :-
$B + 3 D \to 4 A + 2 C$

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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

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 can be:

Q. The rate of reaction for certain reaction is expressed as:

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

The reaction is:

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

Consider the chemical reaction

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

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The rate of reaction is expressed as rate =12d[C]dt=−13d[D]dt=14d[A]dt=−d[B]dt. The reaction is

given Rate expression is 12d[c]dt=−13d[D]dt=14d[A]dt=−d[B]dtReaction is :-B+3D→4A+2C

The rate of reaction is expressed as rate
$= \frac{1}{2} \frac{d [C]}{d t} = - \frac{1}{3} \frac{d [D]}{d t} = \frac{1}{4} \frac{d [A]}{d t} = - \frac{d [B]}{d t}$
. The reaction is

given Rate expression is
$\frac{1}{2} \frac{d [c]}{d t} = \frac{- 1}{3} \frac{d [D]}{d t} = \frac{1}{4} \frac{d [A]}{d t} = \frac{- d [B]}{d t}$
Reaction is :-
$B + 3 D \to 4 A + 2 C$