Question

2A + B = C + D
In this reaction, if we double the concentration of A, reaction rate become 2 times. And in other experiment, we double the concentration of A and B , reaction rate again become two times. What is the order of this reaction ?
a) 1 ,
b) 3 ,
c) 2 ,
d) 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.

For the given reaction : $2A+B\rightarrow C+D$

$Rate=k[A]^x[B]^y$    (1)

k= rate constant

x = order with respect to A

y = order with respect to B

n = x+y = Total order

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

Putting y = 0 in (1) and (2) , we get:

$Rate=k[A]^x[B]^0$  (1)

$2\times Rate=k[2A]^x[B]^0$   (2)

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

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

order of the reaction should be 1 . Bcz on doubling the concentration the rate also doubles..