Notifications
Clear all
0
26/01/2022 5:57 pm
Topic starter
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
1 Answer
0
26/01/2022 6:02 pm
Let the concentration of the reactant be [A] = a
Rate of reaction, R = k [A]2 = ka2
(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
R' = k(2a)2
= 4ka2
= 4R
Therefore, the rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half, i.e. [A] = \(\frac{1}{2}\)a, then the rate of the reaction would be
R' = k\((\frac{1}{2}a)^2\)
= \(\frac{1}{4}ka^2\)
= \(\frac{1}{4}R\)
Therefore, the rate of the reaction would be reduced to \(\frac{1}{4}^{th}\).