The standard free energy change of a reaction is ∆G&de

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 Multiple Choice QuestionsMultiple Choice Questions

71.

20 mL of 0.2 M NaOH is added to 50 mL of 0.2 M acetic acid. The pH of this solution after mixing is (Ka=1.8x 10°)

  • 4.5

  • 2.3

  • 3.8

  • 4


72.

A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. The value of K if the total pressure at equilibrium is 0.8 atm, is

  • 1.8 atm

  • 3 atm

  • 0.3 atm

  • 0.18 atm


73.

For the reaction 2A + B C, H = 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


74.

For the reaction,

2A(g) + B2(g)  2AB2(g) the equilibrium constant, KP at 300K is 16.0. The value of KP for

AB2(g)  A(g) + 1/2B2(g) is

  • 8

  • 0.25

  • 0.125

  • 32


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75.

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

  • K= 102

  • K = 10

  • K = 10-2

  • K = 1


76.

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


77.

The equilibrium constant for the given reaction is 100.

N2(g) + 2O2(g)  2NO2(g)

What is the equilibrium constant for the reaction given below? 

NO2(g) 12 N2(g) + O2(g)

  • 10

  • 1

  • 0.1

  • 0.01


78.

20 mL of 0.1 M acetic acid is mixed with 50 mL of potassium acetate. Ka 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


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79.

The standard free energy change of a reaction is G°= -115 kJ at 298 K. Calculate the equilibrium constant Kp in log Kp (R =8.314 JK -1mol-1 ).

  • 20.16

  • 2.303

  • 2.016

  • 13.83


A.

20.16

G°= -115 × 103 J,T = 298 K, R = 8.314 JK-1 mol-1-G°= 2.303RT log10Kp-(-115 × 103 )= 2.303×8.314× 298 log10Kplog10Kp=1150002.303×8.314× 298log10Kp= 20.16


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80.

Given the equilibrium system:

NH4Cl(s)  NH4+(aq) + Cl-(aq)       (H =+3.5 kcal/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


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