1

### AIPMT 2010 Mains

The reaction,
2A(g) + B(g) $\rightleftharpoons$ 3C(g) + D(g)
is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measuread and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression
A
[(0.75)3 (0.25)] $\div$ [(1.00)2 (1.00)]
B
[(0.75)3 (0.25)] $\div$ [(0.50)2 (0.75)]
C
[(0.75)3 (0.25)] $\div$ [(0.50)2 (0.25)]
D
[(0.75)3 (0.25)] $\div$ [(0.75)2 (0.25)]

## Explanation

2A(g) + B(g) &#8652; 3C(g) + D(g)
Initial mole 1 1 0 0
At equilibrium 1 - (2 $\times$ 0.25)
= 0.5
1 - 0.25
= 0.75
3 $\times$ 0.25
= 0.75
0.25

Equilibrium constant, K = ${{{{\left[ C \right]}^3}\left[ D \right]} \over {{{\left[ A \right]}^2}\left[ B \right]}}$

$\therefore$ K = ${{{{\left[ {0.75} \right]}^3}\left[ {0.25} \right]} \over {{{\left[ {0.5} \right]}^2}\left[ {0.75} \right]}}$
2

### AIPMT 2010 Prelims

In which of the following equilibrium Kc and Kp are not equal?
A
2NO(g) $\rightleftharpoons$ N2(g) + O2(g)
B
SO2(g) + NO2(g) $\rightleftharpoons$ SO3(g) + NO(g)
C
H2(g) + I2(g) $\rightleftharpoons$ 2HI(g)
D
2C(s) + O2(g) $\rightleftharpoons$ 2CO2(g)

## Explanation

As we know, Kp = Kc × (RT)$\Delta$ng

So, for reaction having same number of gaseous moles on reactants and products side will have same value of Kc and Kp otherwise their values are different.

For reaction, 2NO(g) $\rightleftharpoons$ N2(g) + O2(g)

$\Delta$ng = 2 – 2 = 0

$\Rightarrow$ Kp = Kc × (RT)0

$\therefore$ Kp = Kc

For reaction, SO2(g) + NO2(g) $\rightleftharpoons$ SO3(g) + NO(g)

$\Delta$ng = 2 – 2 = 0

$\Rightarrow$ Kp = Kc × (RT)0

$\therefore$ Kp = Kc

For reaction, H2(g) + I2(g) $\rightleftharpoons$ 2HI(g)

$\Delta$ng = 2 – 2 = 0

$\Rightarrow$ Kp = Kc × (RT)0

$\therefore$ Kp = Kc

For reaction, H2(g) + I2(g) $\rightleftharpoons$ 2HI(g)

$\Delta$ng = 2 – 2 = 0

$\Rightarrow$ Kp = Kc × (RT)0

$\therefore$ Kp = Kc

For reaction, 2C(s) + O2(g) $\rightleftharpoons$ 2CO2(g)

$\Delta$ng = 2 – 3 = -1

$\Rightarrow$ Kp = Kc × (RT)-1

$\therefore$ Kp $\ne$ Kc
3

### AIPMT 2010 Prelims

What is [H+] in mol/L of a solution that is 0.20 M in CH3COONa and 0.10 M in CH3COOH? Ka for CH3COOH = 1.8 $\times$ 10$-$5
A
3.5 $\times$ 10$-$4
B
1.1 $\times$ 10$-$5
C
1.8 $\times$ 10$-$5
D
9.0 $\times$ 10$-$6

## Explanation

CH3COOH and CH3COONa constitute to form an acidic buffer.

$\Rightarrow$ pH = pKa + log${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$

pH = –log(1.8 × 10–5) + log ${{\left( {0.20} \right)} \over {\left( {0.10} \right)}}$

= 4.74 + log 2

= 4.74 + 0.3010 = 5.041

Now, pH = – log[H+]

$\Rightarrow$ 5.041 = – log[H+]

$\Rightarrow$ [H+] = 10–5.041 = 9.0 × 106 mol L–1
4

### AIPMT 2010 Prelims

In a buffer solution containing equal concentration of B$-$ and HB, the Kb for B$-$ is 10$-$10. The pH of buffer solution is
A
10
B
7
C
6
D
4

## Explanation

pOH = pKb + log ${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$

$\Rightarrow$ pOH = - log Kb + log ${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$

$\Rightarrow$ pOH = –log10–10 + log 1

[As conc. of HB and B are same]

$\Rightarrow$ pOH = 10

$\Rightarrow$ pH = 14 – pOH = 14 – 10 = 4