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**Problem. The carbon-14 activity of an old wood sample is found to be 14.2 disintegration min ^{â€“1} g^{â€“1}. Calculate age of old wood sample, if for a fresh wood sample carbon-14 activity is 15.3 disintegration min^{â€“1} g^{â€“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 â€¦(1)

Activity of new wood = k N_{new }= 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

The correct answer is (4).

**Problem. Using cuvettes of 0.5 cm path length, a 10 ^{â€“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 ^{â€“1} cm^{â€“1} (2) 3010 M^{â€“1} cm^{â€“1} (3) 5000 M^{â€“1} cm^{â€“1 }(4) 6020 M^{â€“1} cm^{â€“1}**

**Sol. **

Transmittance = T

Absorbance = A =

âˆˆ= 6020 M^{â€“1} cm^{â€“1}

The correct answer is (4).

**Problem. The rate law for one of the mechanisms of the pyrolysis of CH _{3}CHO at 520Â°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 [CH_{3}CHO]^{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 k _{3} and k(Michaelis constant, mol L^{â€“1)} are **

**(1) 1.4 Ã— 10 ^{12}, 10^{4} (2)1.4 Ã— 10^{8}, 10^{4} (3) 1.4 Ã— 10^{8}, 10^{â€“4} (4) 1.4 Ã— 10^{12}, 10^{â€“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

k_{3} = 1.4 Ã— 10^{12} Ã— 10^{â€“4 }

= 1.4 Ã— 10^{8}

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 A _{2 }undergoing dissociative adsorption is **

**(1) **** (2) **** (3) **** (4)**

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

then

if R_{2}(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) k _{1}k_{2}k_{3}[A]^{3}[B] **

**(2) k _{1}k_{2}k_{3}[A][B]^{3} **

**(3) k _{1}k_{2}k_{3}[A][B]^{2} **

**(4) k _{1}k_{2}k_{3}[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 ^{19}Ne and ^{14}C emit a position and b-particle respectively. The resulting species formed are respectively**

**(1) ^{19}Na and ^{14}B (2) ^{19}F and ^{14}N (3) ^{19}Na and ^{14}N (4) ^{19}F and ^{14}B **

**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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