For a reaction between A and B the order with respect to A is 2 and the other with respect to B is 3. The concentrations of both A and B are doubled, the rate will increase by a factor of
A
12
B
16
C
32
D
10
Explanation
Rate1 = k[A]2[B]3
when concentrations of both A and B are
doubled then
Rate2 = k[2A]2[2B]3 = 32 k[A]2[B]3
$$ \therefore $$ Rate will increase by a factor of 32.
3
AIPMT 2012 Prelims
MCQ (Single Correct Answer)
In a zero-order reaction, for every 10oC rise of temperature, the rate is doubled. If the temperature is increased from 10oC to 100oC, the rate of the reaction will become
A
256 times
B
512 times
C
64 times
D
128 times
Explanation
For energy 10° rise in temperature the rate of
reaction doubles. So, rate = 2n when, n = 1 rate = 21
= 2
when, temperature is increased from 10°C to 100°C
change in temperature = 100 – 10 = 90°C
i.e., n = 9
So, rate = 29
= 512 times
4
AIPMT 2012 Prelims
MCQ (Single Correct Answer)
In a reaction, A + B $$ \to $$ product, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentration of both the reactants (A and B) are doubled, rate law for the reaction can be written as
A
rate = k[A][B]2
B
rate = k[A]2[B]2
C
rate = k[A][B]
D
rate = k[A]2[B]
Explanation
Rate of reaction for A + B $$ \to $$ Product
Rate = k[A]x[B]y …(1)
where, x and y are order w.r.t. A and B respectively.
When the concentration of only B is doubled, the
rate is doubled, so
R’ = k [A]x
[2B]y
= 2R …(2)
If concentration of both the reactants A and B are
doubled then the rate increases by a factor of 8 so
R’’ = k[2A]x[2B]y
= 8R ...(3)
= k2x
2y
[A]x
[B]y
= 8R …(4)
From equation (1) and (2), we get
$${{2R} \over R} = {{{{\left[ A \right]}^x}{{\left[ {2B} \right]}^y}} \over {{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}$$
$$ \Rightarrow $$ 2 = 2y
$$ \Rightarrow $$ y = 1
From equation (1) and (4), we get
$${{8R} \over R} = {{{2^x}{2^y}{{\left[ A \right]}^x}{{\left[ B \right]}^y}} \over {{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}$$
$$ \Rightarrow $$ 8 = $${{2^x}{2^y}}$$
Substituting the value of y gives
$${{2^x}{2^1}}$$ = 8
$$ \Rightarrow $$ $${{2^x}}$$ = 4
$$ \Rightarrow $$ x = 2
By replacing the values of x and y in
rate law; rate = k[A]2[B]
Questions Asked from Chemical Kinetics
On those following papers in MCQ (Single Correct Answer)
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