AIPMT 2009

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MCQ (Single Correct Answer)

In the reaction,
BrO$$_{3(aq)}^ - $$ + 5Br$$_{(aq)}^ - $$ + 6H⁺ $$ \to $$ 3Br_2(l) + 3H₂O_(l).
The rate of appearance of bromine (Br₂) is related to rate of disappearance of bromide ions as

$${{d\left[ {B{r_2}} \right]} \over {dt}} = - {5 \over 3}{{d\left[ {B{r^ - }} \right]} \over {dt}}$$

$${{d\left[ {B{r_2}} \right]} \over {dt}} = {5 \over 3}{{d\left[ {B{r^ - }} \right]} \over {dt}}$$

$${{d\left[ {B{r_2}} \right]} \over {dt}} = {3 \over 5}{{d\left[ {B{r^ - }} \right]} \over {dt}}$$

$${{d\left[ {B{r_2}} \right]} \over {dt}} = - {3 \over 5}{{d\left[ {B{r^ - }} \right]} \over {dt}}$$

Explanation

Rate = $${1 \over 3}{{d\left[ {B{r_2}} \right]} \over {dt}} = - {1 \over 5}{{d\left[ {B{r^ - }} \right]} \over {dt}}$$

$$ \therefore $$ $${{d\left[ {B{r_2}} \right]} \over {dt}} = - {3 \over 5}{{d\left[ {B{r^ - }} \right]} \over {dt}}$$

AIPMT 2009

MCQ (Single Correct Answer)

For the reaction, N₂ + 3H₂ $$ \to $$ 2NH₃, if
$${{d\left[ {N{H_3}} \right]} \over {dt}}$$ = 2 $$ \times $$ 10^$$-$$4 mol L^$$-$$1 s^$$-$$1,
the value of $${{ - d\left[ {{H_2}} \right]} \over {dt}}$$ would be

4 $$ \times $$ 10^$$-$$4 mol L^$$-$$1 s^$$-$$1

6 $$ \times $$ 10^$$-$$4 mol L^$$-$$1 s^$$-$$1

1 $$ \times $$ 10^$$-$$4 mol L^$$-$$1 s^$$-$$1

3 $$ \times $$ 10^$$-$$4 mol L^$$-$$1 s^$$-$$1

Explanation

N₂ + 3H₂ $$ \to $$ 2NH₃

$$ - {1 \over 3}{{d\left[ {{H_2}} \right]} \over {dt}} = {1 \over 2}{{d\left[ {N{H_3}} \right]} \over {dt}}$$

$$ \Rightarrow $$ $$ - {{d\left[ {{H_2}} \right]} \over {dt}} = {3 \over 2}{{d\left[ {N{H_3}} \right]} \over {dt}}$$

= $${3 \over 2} \times 2 \times {10^{ - 4}}$$

= $$3 \times {10^{ - 4}}$$ mol L^$$-$$1 s^$$-$$1

AIPMT 2008

MCQ (Single Correct Answer)

The bromination of acetone that occurs in acid solution is represented by this equation.
CH₃COCH_3(aq) + Br_2(aq) $$ \to $$
CH₃COCH₂Br(aq) + H⁺_(aq) + Br^$$-$$_(aq)
These kinetic data were obtained for given reaction concentrations.

Initial concentrations, M
[CH₃COCH₃	[Br₂]	[H⁺]
0.30	0.05	0.05
0.30	0.10	0.05
0.30	0.10	0.10
0.40	0.05	0.20

Initial rate, disappearance of Br₂, Ms^$$-$$1
			5.7$$ \times $$10^$$-$$5
			5.7$$ \times $$10^$$-$$5
			1.2$$ \times $$10^$$-$$4
			3.1$$ \times $$10^$$-$$4

Based on these data, the rate equation is

Rate = k[CH₃COCH₃][Br₂][H⁺]²

Rate = k[CH₃COCH₃][Br₂][H⁺]

Rate = k[CH₃COCH₃][H⁺]

Rate = k[CH₃COCH₃][Br₂]

Explanation

From the first two experiments, it is clear that when concentration of Br₂ is doubled, the initial rate of disappearance of Br₂ remains unaltered. So, order of reaction with respect to Br₂ is zero. The probable rate law for the reaction will be : k[CH₃COCH₃ ][H⁺].

AIPMT 2008

MCQ (Single Correct Answer)

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

2000 K

$${{1000} \over {2.303}}K$$

1000 K

$${{2000} \over {2.303}}K$$

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 = $${{1000} \over {2.303}}K$$