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# Examples - Photochemistry Chemistry Notes | EduRev

## Chemistry : Examples - Photochemistry Chemistry Notes | EduRev

The document Examples - Photochemistry Chemistry Notes | EduRev is a part of the Chemistry Course Physical Chemistry.
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Problem. The carbon-14 activity of an old wood sample is found to be 14.2 disintegration minâ€“1 gâ€“1. Calculate age of old wood sample, if for a fresh wood sample carbon-14 activity is 15.3 disintegration minâ€“1 gâ€“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 Nold = 14.2 â€¦(1)
Activity of new wood = k Nnew = 15.3 â€¦(2)

From equation (1) & (2) we get

or â€¦(3)

We know that k Ã— t1/2 = 0.693

â€¦(4)

We know that

i.e.

T = 617 year

Problem. Using cuvettes of 0.5 cm path length, a 10â€“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 Mâ€“1 cmâ€“1 (2) 3010 Mâ€“1 cmâ€“1 (3) 5000 Mâ€“1 cmâ€“1 (4) 6020 Mâ€“1 cmâ€“1

Sol.

Transmittance = T

Absorbance = A =

âˆˆ= 6020 Mâ€“1 cmâ€“1

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

The overall activation energy Ein 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

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

The value of k3 and k(Michaelis constant, mol Lâ€“1) are

(1) 1.4 Ã— 1012, 104 (2)1.4 Ã— 108, 104 (3) 1.4 Ã— 108, 10â€“4 (4) 1.4 Ã— 1012, 10â€“4

Sol.  We know that Michaelis Menten equation is:

â€¦(1)

Multiply this equation by

or                                         â€¦(2)

and                                       â€¦(3)

Comparing equation (2) & (3) we get

i.e.
and

k3 = 1.4 Ã— 1012 Ã— 10â€“4

= 1.4 Ã— 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 Aundergoing 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)

thenâ€¦(1)

(fast equilibrium)

then       â€¦(2)
(slow)

The rate of formation of product P is

â€¦(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 14

Sol.

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

Problem. The half life of a zero order reaction (A â†’ 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).

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## Physical Chemistry

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