Question

In a reaction \[\ce{mA + nB -> products}\], when the concentration of A is doubled, the rate of reaction is also doubled. When the concentration of B is doubled, the rate of reaction becomes four times. Find the overall order of reaction.

Answer 1

The given reaction is \[\ce{mA + nB -> products}\]

Let the rate law be:

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

Where:

x = order of reaction w.r.t. A

y = order of reaction w.r.t. B

x + y = overall order of the reaction

When [A] is doubled:

Rate ∝ [A]^x

`"New rate"/"Original rate" = ((2[A])/([A]))^x` = 2^x

But it is given that rate doubles, so:

2^x = 2

⇒ x = 1

When [B] is doubled:

Rate ∝ [B]^y

`"New rate"/"Original rate" = ((2[B])/([B]))^y` = 2^y

But it is given that the rate becomes 4 times, so:

2^y = 4

⇒ y = 2

∴ Overall Order = x + y

= 1 + 2

= 3