Test: Rate of a Chemical Reaction and factors influencing rate of reaction(6 Nov) - JEE MCQ

Let rate law be 

1 x 10^-4 = K(0.01)ⁿ

1.41 x 10^-4 = K(0.02)ⁿ

∴ 1.41 = (2)ⁿ 

(2)^1/2 = (2)ⁿ   (Since, √2 = 1.41)

∴ n= 1/2

Thus rate law is 

Rate = k (A)[B]

The given reaction is first order in A and first order is B.
Thus, total order = 2

(a) Unit of k = cone ^{1 - n} time ^-1 = conc^-1 time^-1 Thus, (a) is false.
(b)  of second-order reaction, thus (b) is false. 

Thus, (c) is correct.

Thus, value of k is dependent on the concentration of A and B. Thus, (d) is false.

t₅₀ is independent of concentration of B then w.r.t. B, (n - 1) = 0. Thus, n = 1
Thus, order w.r.t. B = 1
Also, if concentration is made m-times, rate becomes, mⁿ-times. 

Hence, w.r.t.A 

m = 2, mⁿ = 2

∴ 2ⁿ = 2 = 2¹

Thus  n = 1
Thus, order w.r.t. A = 1
Thus, total order = 2 = n
Unit of n th order reaction = conc^(1-n) time^-1
conc in mol L^-1 and time in seconds.

Then, 

(a)

I, II, V - represent first order equation

III, VI - zeroth order equation

IV - second order equation

For n^th order reaction (n > 1)

This equation is true, if n = 2

n = order of the reaction =slope of the line
Larger the value of the slope of the line, larger the order.
Thus, P = 3, Q = 2, R = 1, S = 0.

Thus, order = zero Rate is independent of concentration of A.
Thus, rate remains constant.

For nth order reaction,

(T₅₀ ) is t_1 ,at  [A] = a₁ and is t₂ at [A] = a₂

∴ 

(2) = (2)^n-1

∴ n - 1 = 1, n = 2
 

T₅₀ is independent of concentration of [A], Thus, it follows first order kinetics.