Solved Examples

Example : 1

The experimental data for the reaction 2A + B2 → 2AB  is as follows: 

Write the most probable equation for the rate of reaction giving reason for your answer. 

Solution :

From an examination of above data, it is clear that when the concentration of  B₂ is doubled, the rate is doubled. Hence the order of reaction with respect to B₂ is one. 
Further when concentration of A is doubled, the rate remain unaltered. So, order of reaction  with respect to A is zero. 
 The probable rate law for the reaction will be -dx/dt = k[B₂][A]⁰  = k[B₂] 
 Alternatively Rate = k[B₂]^α
 1.6 × 10-4 = k[0.5]^α
 3.2 × 10-4 = k[1]^α
 On dividing we get α = 1 
 .·. Rate = k[A]⁰ [B₂]   = k[B₂] 

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Example : 2

For the reaction   A + 2B → 2C  the following data were obtained. 

Write down the rate law for the reaction. 

Solution :

Let the rate law be 
-dx/dt = k[A]^x  [B]^y
By keeping the concentration of B constant in experiments (1), (2) and (3) and increasing  concentration uniformly, the rate also increases uniformly, thus, 
Rate ∞ [A], i.e., x = 1 
By keeping the concentration of A constant in experiments (1), (4) and (5) and increasing  the concentration of B, the rate remains the same. 
Hence, y = 0 
The rate law is -dx/dt = k[A] 

Alternatively method:
From Expt. (1), k[1.0]^x [1.0]^y  = 0.15 ....(i) 
From Expt. (2), k[2.0]^x [1.0]^y = 0.30 ... (ii) 
Dividing Eq. (ii) by Eq. (i),  [2.0]^x /[1.0]^x  = 0.30/0.15 = 2 

So , x = 1 

From Expt. (1), k[1.0]^x [1.0]^y = 0.15 ....(i) 
From Expt. (4), k[1.0]^x[2.0]^y  = 0.15 .... (iii) 
Dividing Eq. (iii) by Eq. (i), 
[2.0]^y /[1.0]^y = 1 
So , y = 0 
Hence, the rate law is -dx/dt = k[A]. 

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Example : 3

For the reaction:  2NO + Cl2 → 2NOCl  at 300 K following data are obtained. 

Write rate of law for the reaction. What is the order of the reaction?  Also calculate the specific rate constant. .

Solution

Let the rate law for the reaction be 

Rate = k[NO]^x [Cl₂]^y

From Expt. (1), 

1.2×10-4 = k[0.010]^x [0.010]^y ... (i) 

From Expt. (2),

2.4×10^-4 = k[0.010]^x[0.020]^y  ... (ii) 

Dividing Eq. (ii) by Eq. (i), 

(9.6×10-4)/(2.4×10-4) = ([0.020]^x  )/[0.010]^x
or 2 = (2)^y
 y = 1 
From Expt. (2), 2.4×10-4 = k[0.010]^x [0.020]^y .... (ii) 
From Expt. (3), 9.6×10-4 = k[0.020]^x [0.020]^y .... (ii) 
Dividing Eq. (ii) by Eq. (ii), 
(9.6 × 10-4)/(2.4 × 10-4) = ([0.020]x )/[0.010]x
or 4 = 2x
x = 2 
Order of reaction = x + y = 2 + 1 = 3 
Rate law for the reaction is 
Rate = k[NO]² [Cl₂] 
Considering Eq. (i) again, 
1.2 × 10^-4 = k[0.010]² [0.010] 
 k = (1.2×10-4)/[0.010]³  = 1.2×10² mol^-2 litre² sec-1
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Example :4

For the hypothetical reaction   2A + B → products  the following data are obtained. 

Find out how the rate of the ration depends upon the concentration of A and B and fill in the 
blanks. 

Solution : 

From Expt. (2) and (3), it is clear that when concentration of A is kept  constant and that of B is doubled, the rate increases four times. This shows that the  reaction is of second order with respect to B. 
Similarly, from Expt. (1) and (2), it is observed that when concentration of A is increased three times and that of B two times, the rate becomes twelve times. Hence, the reaction is 
first order with respect to A. 
Thus the rate law for the reaction is 
 Rate = k[A][B]2

(iv) 

(v)

From Expt. (1), 1.75×10-4 = k[0.0017]^x [0.0017]^y [1.0]^z
From Expt. (4), 3.50×10-4 = k[0.0017]^x [0.0017]^y [0.5]^z
 (1.75 × 10-4)/(3.50 × 10-4) = [1.0]^z /[0.5]^z
or 1/2 = 2z
or 2-1 = 2z
z = -1, i.e., order w.r.to OH^-  is -1. 
Rate law = k[OCI- ][I- ]/[OH-] 
From Expt.(1)k = (1.75×10^-4[OH^- ])/[OCI^-][I^-] = (1.75 × 10^-4×1.0)/(0.0017×0.0017)  = 60.55 s^-1
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Example : 5

The rate law for the reaction,   2Cl₂O → 2Cl₂ + O₂ at 200oC is found to be : rate = k[Cl₂O]²
(a) How would the rate change if [Cl₂O] is reduced to one-third of its original value? 
(b) How should the [Cl₂O] be changed in order to double the rate? 
(c) How would the rate change if [Cl₂O] is raised to threefold of its original value? 

Solution : 

(a) Rate equation for the reaction, 
 r = k[Cl₂O]²
 Let the new rate be r'; so 
 r' = k[(Cl₂ O)/3]²  = 1/9 r 
(b) In order to have the rate = 2r, let the concentration of Cl₂O be x. 
 So 2r = kx²  .... (i) 
 We know that r = k[Cl₂O]²  .... (ii) 
 Dividing Eq. (i) by (ii), 
 2r/r = (kx² )/(k[Cl₂O]₂ ) 
 or 2 = x²/[Cl₂O]₂
 or x²  = 2[Cl₂O]₂
 or x = √√2 [Cl₂ O] 
(d) New rate = k[3Cl₂O]²  = 9k[Cl₂O]² = 9r
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Example : 6

For a reaction in which A and B from C, the following data were obtained from  three experiments: 

What is the rate equation of the reaction and what is the value of rate constant? 

Solution: 

Let the rate equation be k[A]^x [B]^y
From Expt. (1), 0.3×10^-4 = k[0.03]^x[0.03]^y ... (i) 
From Expt. (2), 1.2×10^-4 = k[0.06]^x [0.06]^y ... (ii) 
(1.2 × 10-4)/(0.3×10-4) = ([0.06]^x [0.06]^y)/([0.03]^x [0.03]^y)  = 2^x × 2y = 4 ... (iii) 
Similarly from Expt. (1) and (3), 
2x ×3y = 9 ....(iv) 
Solving Eq. (iii) and (iv), 
x = 0, y = 2 
Rate equation, Rate = k[B]²
Considering Eq. (i) again, 
 k = (0.3×10^-4)/[0.03]² = 3.33 × 10-2 mol L^-1 s^-1
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Example : 7

For a first order reaction when log k was plotted against , a straight 1/T line with a slope of -6000 was obtained. Calculate the activation energy of the reaction 

Solution

Slope = -Ea/2.303R = –6000

Ea = 6000  2.303 8.314

= 1.148 105J

 = 27483.935Cal  = 27.48 Kcal

Example : 8

For a reaction, the energy of activation is zero. What is the value of rate constant at 300 K if k = 1.6  106 s^–1 at 280 K? (R = 8.31 J K^–1 mol^–1)

Solution:             

We know,

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Example : 9

The time required for 10% completion of first order reaction at 298 K is equal to that required for its 25% completion at 308K. If the preexponential factor for the reaction is 3.56 ´ 10⁹ s^–1, calculate the energy of activation

Solution:

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Question : 10

At 380°C, the half-life period for the first order decomposition of H₂O₂ is 360 min. the energy of activation of the reaction is 200 kJ mol^–1, Calculate the time required for 75% decomposition at 450°C.

Solution:

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