1
MCQ (Single Correct Answer)

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) ⇌ 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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

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