For the reaction $2\mathrm{A} + \mathrm{B}\rightleftharpoons \mathrm{C}, ∆\mathrm{H} = \mathrm{x}$ cal, which one of the following conditions would favour the yield of C on the basis of Le-Chatelier principle?

• High pressure, high temperature

• Only low temperature

• High pressure, low temperature

• Only low pressure

For the reaction,

$2\mathrm{A}\left(\mathrm{g}\right) + {\mathrm{B}}_2\left(\mathrm{g}\right) \rightleftharpoons 2{\mathrm{AB}}_2\left(\mathrm{g}\right)$ the equilibrium constant, K_P at 300K is 16.0. The value of K_P for

${\mathrm{AB}}_2\left(\mathrm{g}\right) \rightleftharpoons \mathrm{A}\left(\mathrm{g}\right) + 1/2{\mathrm{B}}_2\left(\mathrm{g}\right) \mathrm{is}$

• 8

• 0.25

• 0.125

• 32

In which of the following case, does the reaction go farthest to completion?

• K= 10²

• K = 10

• K = 10^-2

• K = 1

74.

pH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27C. The buffer capacity of the solution is

• 0.1

• 10

• 1

• 0.4

The equilibrium constant for the given reaction is 100.

${\mathrm{N}}_2\left(\mathrm{g}\right) + 2{\mathrm{O}}_2\left(\mathrm{g}\right) \rightleftharpoons 2{\mathrm{NO}}_2\left(\mathrm{g}\right)$

What is the equilibrium constant for the reaction given below? 

${\mathrm{NO}}_2\left(\mathrm{g}\right) \rightleftharpoons \frac{1}{2} {\mathrm{N}}_2\left(\mathrm{g}\right) + {\mathrm{O}}_2\left(\mathrm{g}\right)$

• 10

• 1

• 0.1

• 0.01

20 mL of 0.1 M acetic acid is mixed with 50 mL of potassium acetate. K_a of acetic acid = 1.8 x 10^-5 at 27C. Calculate concentration of potassium acetate if pH of the mixture is 4.8.

• 0.1 M

• 0.04 M

• 0.4 M

• 0.02 M

The standard free energy change of a reaction is $∆\mathrm{G}°= -115 \mathrm{kJ} \mathrm{at} 298 \mathrm{K}.$ Calculate the equilibrium constant K_p in log K_p (R =8.314 JK ^-1mol^-1 ).

• 20.16

• 2.303

• 2.016

• 13.83

Given the equilibrium system:

${\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{s}\right) \to {\mathrm{NH}}_4^{+}\left(\mathrm{aq}\right) + {\mathrm{Cl}}^{-}\left(\mathrm{aq}\right) (∆\mathrm{H} =+3.5 \mathrm{kcal}/\mathrm{mol}).$

What change will shift the equilibrium to the right?

• Decreasing the temperature

• Increasing the temperature

• Dissolving NaCl crystals in the equilibrium mixture

• Dissolving NH₄NO₃ crystals in the equilibrium mixture

Equivalent amounts of H₂ and I₂, are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to HI, the K_c at this temperature is :

• 64

• 16

• 0.25

• 4

For the reaction ${\mathrm{H}}_2\left(\mathrm{g}\right) + {\mathrm{I}}_2\left(\mathrm{g}\right) \to 2\mathrm{HI}\left(\mathrm{g}\right),$ the equilibrium constant K_P changes with: 

• total pressure

• catalyst

• the amount H₂ and I₂

• temperature