Chemistry, asked by ruhuamin123, 11 months ago

5.
In an elementary reaction A + 2B + 2C + D. If
the concentration of A is increased four times and B
is decreased to half of its initial concentration then
the rate becomes
(1) Twice
(2) Half
(3) Unchanged
(4) One fourth of the rate​

Answers

Answered by kobenhavn
35

Answer: (3) Unchanged

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

For elementary reactions, the stoichiometry gives the order of the reaction.

Rate=k[A]^x[B]^y

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

Rate law=R=k[A]^1[B]^2    (1)

a) When concentration of A is increased four times and B  is decreased to half of its initial concentration

R'=k[4A]^1[\frac{B}{2}]^2

R'=k[4]^1[A]^1[\frac{1}{2}]^2[B}^2    (2)

Dividing 2 by 1 :\frac{R'}{R}=\frac{k[4]^1[A]^1[\frac{1}{2}]^2[B]^2}{k[A]^1[B]^2}

\frac{R'}{R}=1

R'=R

Thus the rate remains unchanged.

Answered by Rithy01
15

Answer:unchanged

Explanation:pls refer the attached pic HOPE IT HELPS YOU

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