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)
⇌
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 = $${{{{\left[ C \right]}^3}\left[ D \right]} \over {{{\left[ A \right]}^2}\left[ B \right]}}$$
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
Questions Asked from Equilibrium
On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
