Test: Determination of Rate of a Reaction - NEET 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.

Cl-atom attached to N-atom, is an oxidising agent and can oxidise Kl to l₂.

l₂ can be titrated using hypo in iodometric titration.

Cl-atom attached to benzene nucleus is not an oxidising agent.

A → B

[B] increases uniformly with time t. Order is zero. Thus, at every point,  = constant

We find that,    t₇₅ = 2 x t₅₀

Thus, given reaction is of first-order.

(i) Rate remains constant

(dx/dt) = k(A)ⁿ = [A]⁰ = k

Thus   order = 0

Thus,(i)→(s)
(ii) Half-life period is independent of concentration

∴  (n-1) = 0

∴  n = 1

Thus, (ii) →(r)

Thus, graph is for second-order reactions

Thus, (iii) → (q)

0 min 5.0 x 10¹⁴  molecules L^-1
30 min 2.5 x 10¹⁴ molecules L^-1
60 min 1.25  x 10¹⁴ molecules L^-1
At 30 min, 50% reaction takes place.
At 60 min, 75% reaction takes place.
Thus,   T₇₅ = 2 x T₅₀
it is valid for first order reaction. Thus, order = 1

Time : [PAN]

0 min 5.0 x 10¹⁴ molecules L^-1

30 min 2.5  x 10¹⁴ molecules L^-1

60 min  2.5 x 10¹⁴ molecules L^-1
At 30 min, 50% reaction takes place.
At 60 min, 75% reaction takes place.

Thus,   T₇₅ = 2 x t₅₀

It is valid for first order reaction. Thus, order = 1

For first order reactions, T₅₀ = 0.693/k

-dp/dt = k(P_NO)^a (P_H2)^b

0.0012 = k(0.25)^a(0.2)^b

0.0024 = k(0.50)^a(0.1)^b

0.0048 = k(0.50)^a(0.2)^b

From Eqs. (i) and (iii), we get 

4 = (2)^a

∴   (2)² = (2)^a

∴   a = 2

Thus w.r.t. No, order = 2

From Eqs.)ii) and (iii), we get 

2 = (2)^b

(2)¹ = (2)^b

∴   b = 1
Thus, w.r.t. H₂, order = 1

Total order = 3

From Eq. (i),

0.0012 =  k(0.25)²(0.2)¹ = k (0.25)²(0.2)

∴  k = 0.0012/(0.25)2(0.02) = 0.096 atm^-2 min^-1

0.0012 = k(0.25)²(0.2)¹ = k(0.25)²(0.2)

∴   k = 0.0012/(0.25)² (0.2) = 0.096 atm^-2 min^-1

where, a == order w.r.t. A, 6 = order w.r.t. B

(i) 5.0 x 10^-3 = k[0.1]^a[1.0]^b

(ii) 2.0 x 1^-2 = k [0.1]^a[2.0]^b

(iii) 2.5 x 10^-3 = k [0.05]^a[1.0]^b

From Eqs. (i) and (ii), we get

Order w.r.t.A = 1

From Eq. (ii), 5.0x10^-3 = K(0.1)(1.0)²

∴  k = 5.0 x 10^-2M^-2h^-1

where, a == order w.r.t. A, 6 = order w.r.t. B

(i) 5.0 x 10^-3 = K [0.1]^a [1.0]^b

(ii) 2.0 x 10^-2 = k [ 0.1]^a[2.0]^b

(iii) 2.5 x 10^-3 = k [0.05]^a[1.0]^b

From Eqs. (i) and (ii), we get

(2)² = (2)^b

∴ b = 2

Order w.r.t.B = 2

From Eqs. (i) and (iii), we get

(2) = (2)^a

a = 1

Order w.r.t.A 

From Eq. (ii), 

5.0 x 10^-3 = k (0.1) (1.0)²

∴ k = 5.0 x 10^-2m^-2h^-1