Home
Class 12
CHEMISTRY
The rate of a reaction starting with ini...

The rate of a reaction starting with initial concentration of `2 xx 10^(-3)` and `1 xx 10^(-3) M` are equal to `2.40 xx 10^(-40)` and `0.60 xx 10^(-4) M s^(-1)`, respectively. Calculate the order or reaction w.r.t. reactant and also the rate constant.

Text Solution

Verified by Experts

`(r_(0))_(1) = k[A_(0)]_(1)^(a)` (`a` is order of reaction)
`(r_(0))_(2) = k[A_(0)]_(2)^(a)`
`((r_(0))_(1))/((r_(0))_(2)) = {([A_(0)]_(1))/([A_(0)]_(2))}^(a)`
or `a = (log(r_(0))_(1) - log(r_(0))_(2))/(log[A_(0)]_(1)-log[A_(0)]_(2))`
`= (log[2.40 xx 10^(-4)]-log [0.60 xx 10^(-4)])/(log[2 xx 10^(-3)]-log[1xx10^(-3)])`
`= (-3.62 + 4.22)/(-2.70 + 3) = 2`
`r = k[A]^(2)`
Also, `k = (r )/([A]^(2)) = (2.40 xx 10^(-4))/([2 xx 10^(-3)]^(2)) = 60 mol^(-1) L s^(-1)`
Promotional Banner

Similar Questions

Explore conceptually related problems

Rate constant for a reaction is 1.85 xx 10^(2)s^(-1) . Give the order of reaction.

Identify the order of the reaction from the rate constant K= 2.3 xx 10^(-6) L " mol"^(-1)s^(-1)

In a reversible reaction the rate constants of the forward and the backward reactions are 4.8xx10^(-5)s^(-1)and1.2xx10^(-4)s^(-1) respectively. Calculate the equilibrium constant.

The rate of the reaction A rarr Products at the initial concentration of 3.24 xx10^(-2) M is nine times its rate at another initial concentration of 1.2xx10^(-3) M. The order of the reaction is

If the rate constant of a reaction is k = 3 xx 10^(-4) s^(-1) , then identify the order of the reaction.

Rate constant of a reaction is k = 3.14 xx 10^(-4) mol L^(-1)s^(-1) . What is the order of the reaction.

The rate constant of a first order reaction at 300 K and 310 K are respectively 1.2 xx 10^(3) s^(-1) and 2.4 xx 10^(3) s^(-1) . Calculate the energy of activation. (R = 8.314 J K^(-1) mol^(-1))