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1

AIPMT 2011 Prelims

MCQ (Single Correct Answer)
The value of $$\Delta $$H for the reaction
X2(g) + 4Y2(g) $$\rightleftharpoons$$ 2XY4(g)
is less than zero. Formation of XY4(g) will be favoured at
A
high temperature and high pressure
B
low pressure and low temperature
C
high temperature and low pressure
D
high pressure and low temperature

Explanation

X2(g) + 4Y2(g) $$\rightleftharpoons$$ 2XY4(g)

$$\Delta $$ng = -ve and $$\Delta $$H = -ve

As $$\Delta $$H < 0 i.e., the given reaction is exothermic. According to Le-Chatelier principle, for exothermic reaction, forward reaction is favoured when temperature becomes low. Also, there are 5 gaseous moles on reactant side and 2 gaseous moles on products side. So, forward reaction is favoured when pressure of the reaction mixture becomes high. The reason is that at high pressure reaction tends to more in direction where there is lessee number of gaseous moles.
2

AIPMT 2010 Mains

MCQ (Single Correct Answer)
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 = $${{{{\left[ {0.75} \right]}^3}\left[ {0.25} \right]} \over {{{\left[ {0.5} \right]}^2}\left[ {0.75} \right]}}$$
3

AIPMT 2010 Prelims

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

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