Chemical Kinetics

Question

2A + 3B ----> Product

Order of reaction wrt A -> 2

Order of reaction wrt B -> -1

(i) Write rate law expression.

(ii) Calculate order of reaction

(iii) What happens when A is doubled alone

(iv) What happens when B is halved alone

Answer 1

2A + 3B → Product

given, order of reaction w.r.to A = 2

order of reaction w.r.to B = -1

(i) rate of law expression

R₁ = rate = \(\frac{k[A]^2}{[B]}\)

(ii) order of reaction = 2 + (-1) = 1

(iii) if concentration A doubled , then

R₂ = \(\frac{k[2A]^2}{[B]}\)

R₂ = \(\frac{k\times4[A]}{[B]}\)

\(\frac{R_2}{R }\) = 4

⇒ R₂ = 4R

On doubleing the concentration of (A) reaction rate become four times.

(iv) if concentration of B become halved

then,

R₃ = \(\frac{k[A]^2}{[\frac{B}{2}]}\)

R₃ = \(\frac{k\times2[A]^2}{[B]}\)

\(\frac{R_3}{R }\) = 2

⇒ R3 = 2R

∴ when concentration of B become halved the reaction rate become doubled.