Question

7. A+B
C+D
In this reaction, if we double the concentration of A, reaction rate
we double the concentration of A. reaction rate become two times. And in other experiment
we double the concentration of A and B, reaction rate again bec
not A and B, reaction rate again become two times. What is the order of this
reaction?
(1) 1
(2) 3
(3) 2
(4) 1.5​

Answer 1

The order of this reaction is 1.

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.

$A+B\rightarrow C+D$

$Rate=k[A]^x[B]^y$    (1)

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: $2\times Rate=k[2A]^x[B]^y$    (2)

From trial 2: $2\times Rate=k[2A]^x[2B]^y$    (3)

Dividing 3 by 2 :$\frac{2\times Rate}{2\times Rate}=\frac{k[2A]^x[2B]^y}{k[2A]^x[B]^y}$

$1=2^y,2^0=2^y$ therefore y=0.

b) $Rate=k[A]^x[B]^y$    (1)

From trial 2: $2\times Rate=k[2A]^x[B]^y$    (3)

Dividing  3 by 1 :$\frac{2\times Rate}{Rate}=\frac{k[2A]^x[B]^y}{k[A]^x[B]^y}$

$2=2^x,2^1=2^x$, x=1

Thus rate law is $Rate=k[A]^1[B]^0$

Thus order with respect to A is 1 , order with respect to B is 0 and total order is 1+0=1

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