Examples: Photochemistry & Beer-Lambert's Law | Physical Chemistry PDF Download

Problem. The carbon-14 activity of an old wood sample is found to be 14.2 disintegration min^–1 g^–1. Calculate age of old wood sample, if for a fresh wood sample carbon-14 activity is 15.3 disintegration min^–1 g^–1 (t1/2 carbon =14) = 5730 year), is

(1) 5000 year (2) 4000 year (3) 877 year (4) 617 year

Sol.     Act ivit y =  

 k = rate constant 
and    NA = no. of atom
Activity of old wood = k N_old = 14.2 …(1)
Activity of new wood = k N_new = 15.3 …(2)

From equation (1) & (2) we get

or …(3)

We know that k × t1/2 = 0.693

 …(4)

We know that

i.e.

T = 617 year

The correct answer is (4).

Problem. Using cuvettes of 0.5 cm path length, a 10^–4 M solution of a chromphone shows 50% transmittance at certain wave length. The molar extinction coefficient of the chromphre at this wave length is (log 2 = 0.3010) 

(1) 1500 M^–1 cm^–1 (2) 3010 M^–1 cm^–1 (3) 5000 M^–1 cm^–1 (4) 6020 M^–1 cm^–1

Sol. 

Transmittance = T 

Absorbance = A = 

∈= 6020 M^–1 cm^–1

The correct answer is (4).

Problem. The rate law for one of the mechanisms of the pyrolysis of CH₃CHO at 520°C and 0.2 bar is:
 Rate =  

The overall activation energy E_a in terms of the rate law is:

(1) Ea(2) + Ea(1) + 2Ea(4)   (2)   (3) 
 (4) 

Sol.    Rate =   

= koverall [CH₃CHO]^3/2
i.e.   

i.e.

and

or 

The correct answer is (3).

Problem. In the Michaelis-Menten mechanism of enzyme kinetics, the expression obtained as

The value of k₃ and k(Michaelis constant, mol L^–1) are 

(1) 1.4 × 10¹², 10⁴ (2)1.4 × 10⁸, 10⁴ (3) 1.4 × 10⁸, 10^–4 (4) 1.4 × 10¹², 10^–4

Sol.  We know that Michaelis Menten equation is:

…(1)

Multiply this equation by 

or                                         …(2) 

and                                       …(3)

Comparing equation (2) & (3) we get

i.e. 
and  

k₃ = 1.4 × 10¹² × 10^–4 

= 1.4 × 10⁸
i.e. The correct answer is (3).

Problem. The Langunier adsorption isotherm is given by   where P is the pressure of the adsorbate gas. The Langmuir adsorption isotherm for a diatomic gas A₂ undergoing dissociative  adsorption is 

(1)    (2)   (3)    (4)

Sol.  R(g) + M(surface)      RM (surface)

then  

if  R₂(g) + 2M (surface)     2M (surface)

i.e. the correct answer is (4).

Problem. The overall rate of following complex reaction

             (fast equilibrium)

          (fast equilibrium)

        (slow)

The steady state approximate would be 

(1) k₁k₂k₃[A]³[B] 

(2) k₁k₂k₃[A][B]³ 

(3) k₁k₂k₃[A][B]² 

(4) k₁k₂k₃[A][B] 

Sol.           (fast equilibrium)

then…(1)

          (fast equilibrium)

then       …(2)
        (slow)

The rate of formation of product P is 

…(3)

From equation (1) & (2) we get

&

then

i.e. the correct answer is (1)

Problem. The species ¹⁹Ne and ¹⁴C emit a position and b-particle respectively. The resulting species formed are respectively

(1) ¹⁹Na and ¹⁴B (2) ¹⁹F and ¹⁴N (3) ¹⁹Na and ¹⁴N (4) ¹⁹F and ¹⁴B 

Sol. 

i.e. the correct answer is (b).
 

Problem. The half life of a zero order reaction (A → P) is given by (k = rate constant)

(1)   (2)    (3)   (4)  

Sol.   

if t = t1/2 then

Thus

i.e. the correct answer is (1).