Question

The transformation of the molecule X into Y follows second order kinetics. If the concentration of X increases to three times, how will it affect the rate of formation of Y?

Answer 1

According to question, r₁ = k[X]²        …(i)
If concentration is increased to three times, then
r₂ = k [3X]²
r₂ = k 9 X²
r₂ = 9k [X]²         …(ii)
Divide eq (ii) by eq. (i)
\({ r_{ 2 } }{ r_{ 1 } } =\frac { 9k[X]^{ 2 } }{ k[X]^{ 2 } } \)
\(\frac { r_{ 2 } }{ r_{ 1 } } =9 \)
r₂ = 9 x r_ı
i.e., if concentration is increased there times, the rate increases 9 times.