1
MCQ (Single Correct Answer)

AIPMT 2008

The rate constants k1 and k2 for two different reactions are 1016 $$ \cdot $$ e$$-$$2000/T and 1015 $$ \cdot $$ e$$-$$1000/T, respectively.
The temperature at which k1 = k2 is
A
2000 K
B
$${{1000} \over {2.303}}K$$
C
1000 K
D
$${{2000} \over {2.303}}K$$

Explanation

k1 = 1016 $${e^{ - {{2000} \over T}}}$$

k2 = 1015 $${e^{ - {{1000} \over T}}}$$

On taking log of both the equations, we get

$$\log {k_1} = 16 - {{2000} \over {2.303T}}$$

$$\log {k_2} = 15 - {{1000} \over {2.303T}}$$

As k1 = k2

$$16 - {{2000} \over {2.303T}}$$ = $$15 - {{1000} \over {2.303T}}$$

$$ \Rightarrow $$ T = $${{1000} \over {2.303}}K$$
2
MCQ (Single Correct Answer)

AIPMT 2007

The reaction of hydrogen and iodine monochloride is given as :
H2(g) + 2ICl(g) $$ \to $$ 2HCl(g) + I2(g)
This reaction is of first order with respect to H2(g) and ICl(g),
following mechanisms were proposed.

Mechanism A :
     H2(g) + 2ICl(g) $$ \to $$ 2HCl(g) + I2(g)
Mechanism B :
     H2(g) + ICl(g) $$ \to $$ HCl(g) + HI(g) ; slow
     HI(g) + ICl(g) $$ \to $$ HCl(g) + I2(g) ; fast

Which of the above mechanism(s) can be consistent with the given information about the reaction?
A
A and B both
B
Neither A nor B
C
A only
D
B only

Explanation

The slow step is the rate determining step and it involves 1 molecule of H2(g) and 1 molecule of ICl(g) . Hence the rate will be,

r = k[H2(g)] [ICl(g)]

$$ \therefore $$ The reaction is 1st order with respect to H2(g) and ICl(g).
3
MCQ (Single Correct Answer)

AIPMT 2007

In a first-order reaction A $$ \to $$ B, if k is rate constant and initial concentration of the reactant A is 0.5 M, then the half-life is
A
$${{\log 2} \over k}$$
B
$${{\log 2} \over {k\sqrt {0.5} }}$$
C
$${{\ln 2} \over k}$$
D
$${{0.693} \over {0.5k}}$$

Explanation

For first order reaction

k = $${{2.303} \over t}\log {a \over {a - x}}$$

at $${t_{1/2}}$$, x = $${a \over 2}$$

$${t_{1/2}}$$ = $${{2.303} \over k}\log {a \over {a - {a \over 2}}}$$

= $${{\ln 2} \over k}$$
4
MCQ (Single Correct Answer)

AIPMT 2007

If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately
(log 4 = 0.60, log 5 = 0.69)
A
45 minutes
B
60 minutes
C
40 minutes
D
50 minutes

Explanation

For a first order reaction,

k = $${{2.303} \over t}\log {a \over {a - x}}$$

k = $${{2.303} \over {60}}\log {{100} \over {40}}$$

= $${{2.303} \over {60}}\log 2.5$$

= 0.0153

Also, $${t_{1/2}}$$ = $${{2.303} \over k}\log {{100} \over {50}}$$

= $${{2.303} \over {0.0153}}\log 2$$

= 45 min.

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