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 Multiple Choice QuestionsMultiple Choice Questions

81.

Find the correct equation

  • E2 - E1 - H2 + H1 = n2RT - n1RT

  • E2 - E1 - H2 - H1 = n2RT - n1RT

  • H2 - H1 - E2 + E1 = n2RT - n1RT

  • H2 - H1 - E2 - E1 = n2RT - n1RT


82.

Assuming enthalpy of combustion of hydrogen at 273 K is -286 kJ and enthalpy of fusion of ice at the same temperature to be +6.0 kJ, calculate enthalpy change during formation of 100 g of ice.

  • +1622 kJ

  • - 1622 kJ

  • +292 kJ

  • - 292 kJ


83.

For which among the following reactions, change in entropy is less than zero?

  • Sublimation of iodine

  • Dissociation of hydrogen

  • Formation of water

  • Thermal decomposition of calcium carbonate


84.

Which among the following is a feature of adiabatic expansion?

  • V < 0

  • U < 0

  • U > 0

  • T = 0


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85.

What is the amount of work done when two moles of ideal gas is compressed from a volume of 1m3 to 10 dm3 at 300 K against a pressure of 100 kPa?

  • 99 kJ

  • -99kJ

  • 114.9 kJ

  • -114.9 kJ


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86.

The work done during combustion of 9 × 10-2 kg of ethane, C2H6 (g) at 300 K is (Given R = 8.314 J deg-1 mol-1, atomic mass C = 12, H= 1).

  • 6.236 kJ

  • -6.236 kJ

  • 18.71 kJ

  • -18.71 kJ


C.

18.71 kJ

Work done in a chemical reaction, 

W = ngRT

where, ng = number of moles of gaseous products - number of moles of gaseous reactants.

Combustion of ethane is-

C2H6 (g) + 72O2 (g) → 2CO2 (g) + 3H2O (l)

ng = 2 - 4.5 = -2.5

W = (2.5 mol) (8.314 JK-1 mol-1) × 300 K

    = 6235.5 kJ = 6.2355 J

Now, 1 mole of C2H6 = 30 gm of C2H6 = 6.2355 kJ

Work done for combustion of 30 gm of C2H= 6.2355 kJ

 Work done for combustion of 90 gm of C2H

6.2355 × 9030 

= 18.7065 kJ = 18.71 kJ


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87.

The first law of thermodynamics for isothermal process is

  • q = -W

  • U = W

  • U = qv

  • U = -qv


88.

Identify the invalid equation

  • H = ΣHproducts - ΣHreactants

  • H = U + pV

  • H°(reaction) = ΣH°(products bonds) - ΣH°(reactant bonds)

  • H = U + nRT


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89.

For the process A (1, 0.05 atm, 32°C) → A (g, 0.05 atm, 32°C)

The correct set of thermodynamic parameters is

  • G = 0 and S = -ve

  • G = 0 and S = +ve

  • G = +ve and S = 0

  • G = -ve and S = 0


90.

The standard enthalpy of formation of H2O (l) and Fe2O3 (s) are respectively -286 kJ mol-1 and -824 kJ mol-1. What is the standard enthalpy change for the following reaction?

Fe2O3 (s) + 3H2 (g) → 3H2O (l) + 2Fe (s)

  • -538 kJ mol-1

  • +538 kJ mol-1

  • -34 kJ mol-1

  • +34 kJ mol-1


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