The rate of the reaction- A-B-C- Products- is given by- - d- A - dt -k- A - 1 - 2 - B - 1 - 3 - C - 1 - 4-The order of the reaction is-

The correct option is B 13 12 #160; #160; #160; dA dt =K[ A ] #160; ^ 1 / 2 [ B ] #160; ^ 1 / 3 [ C ] #160; ^ 1 / 4 The order of re ...

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Byju's Answer

Standard XII

Chemistry

Order of Reaction

The rate of t...

Question

The rate of the reaction, $A + B + C ⟶$ Products, is given by, $- \frac{d [A]}{d t} = k {[A]}^{1 / 2} {[B]}^{1 / 3} {[C]}^{1 / 4}$ .
The order of the reaction is:

\frac{1}{2}

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\frac{13}{12}

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Solution

The correct option is B $\frac{13}{12}$
$\frac{- d A}{d t} = K {[A]}^{1 / 2} {[B]}^{1 / 3} {[C]}^{1 / 4}$
The order of reaction is given as sum oof all the partial order.
hence order $= \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{12}$

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Similar questions

The rate of certain hypothetical reaction A + B + C $\to$ Products, is given by $r = - \frac{d A}{d t} = k [A]^{1 / 2} [B]^{1 / 3} [C]^{1 / 4}$ The order of a reaction is given by:

Q. The rate of the reaction,

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is given by:

r = - \frac{d [A]}{d t} = k [A]^{1 / 2} [B]^{1 / 2} [C]^{1 / 4}

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Q. The rate of certain hypothetical reaction

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r = - \frac{d [A]}{d t} = K [A]^{1 / 2} K [B]^{1 / 3} K [C]^{1 / 4}

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For chemical reaction A

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\frac{{- d C}_{A}}{d t} = \frac{K_{1} C_{A}}{1 + K_{2} C_{A}}

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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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The rate of the reaction, A+B+C⟶ Products, is given by, −d[A]dt=k[A]1/2[B]1/3[C]1/4.The order of the reaction is:

The correct option is B 1312 −dAdt=K[A]1/2[B]1/3[C]1/4The order of reaction is given as sum oof all the partial order.hence order =12+13+14=1312

The rate of the reaction, $A + B + C ⟶$ Products, is given by, $- \frac{d [A]}{d t} = k {[A]}^{1 / 2} {[B]}^{1 / 3} {[C]}^{1 / 4}$ .
The order of the reaction is:

The correct option is B $\frac{13}{12}$
$\frac{- d A}{d t} = K {[A]}^{1 / 2} {[B]}^{1 / 3} {[C]}^{1 / 4}$
The order of reaction is given as sum oof all the partial order.
hence order $= \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{12}$