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1

### AIPMT 2012 Mains

Given that the equilibrium constant for the reaction,
2SO2(g) + O2(g) $$\rightleftharpoons$$ 2SO3(g)
has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature ?
SO3(g) $$\rightleftharpoons$$ SO2(g) + $${1 \over 2}$$ O2(g)
A
1.8 $$\times$$ 10$$-$$3
B
3.6 $$\times$$ 10$$-$$3
C
6.0 $$\times$$ 10$$-$$2
D
1.3 $$\times$$ 10$$-$$5

## Explanation

2SO2(g) + O2(g) $$\rightleftharpoons$$ 2SO3(g),    K

SO3(g) $$\rightleftharpoons$$ SO2(g) + $${1 \over 2}$$ O2(g),     K' = $$\sqrt {{1 \over K}}$$

$$\therefore$$ K' = $$\sqrt {{1 \over {278}}} =$$ 6.0 $$\times$$ 10$$-$$2
2

### AIPMT 2012 Mains

Given the reaction between 2 gases represented by A2 and B2 to give the compound AB(g),

A2(g) + B2(g) $$\rightleftharpoons$$ 2AB(g)

At equilibrium, the concentration of
A2 = 3.0 $$\times$$ 10$$-$$3 M, of B2 = 4.2 $$\times$$ 10$$-$$3 M, of AB = 2.8 $$\times$$ 10$$-$$3 M
If the reaction takes place in a sealed vessel at 527oC, then the value of Kc will be
A
2.0
B
1.9
C
0.62
D
4.5

## Explanation

A2(g) + B2(g) $$\rightleftharpoons$$ 2AB(g);

Kc = $${{{{\left[ {AB} \right]}^2}} \over {\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}$$

= $${{{{\left( {2.8 \times {{10}^{ - 3}}} \right)}^2}} \over {3 \times {{10}^{ - 3}} \times 4.2 \times {{10}^{ - 3}}}}$$

= 0.62
3

### AIPMT 2012 Prelims

Equimolar solutions of the following substances were prepared separately. Which one of these will record the highest pH value ?
A
BaCl2
B
AlCl3
C
LiCl
D
BeCl2

## Explanation

All of the given salts have same anion i.e., Cl which on hydrolysis gives HCl which is a strong acid.

Now, among the salts which have cation that gives a strongest base on hydrolysis of salt have the highest pH value. As Ba form Ba(OH)2 which is a stronger base thus, it results in the highest pH value.
4

### AIPMT 2012 Prelims

pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product (Ksp) of Ba(OH)2 is
A
3.3 $$\times$$ 10$$-$$7
B
5.0 $$\times$$ 10$$-$$7
C
4.0 $$\times$$ 10$$-$$6
D
5.0$$\times$$ 10$$-$$6

## Explanation

Ba(OH)2 Ba2+ + 2OH
At equilibrium x 2x

pH = – log[H+]

12 = – log [H+]

$$\Rightarrow$$ [H+] = 10–12

As, [H+][OH ] = 10–14

10–12 [OH ] = 10–14

$$\Rightarrow$$ [OH ] = 10–2

As [OH ] = 2x = 10–2 then x = 5.0 × 10–3

Now, Ksp = [Ba2+][OH ]2

Ksp = (5 × 10–3) (10–2)2 = 5.0 × 10–7

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