Hess's law is based on : from Chemistry Thermodynamics

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 Multiple Choice QuestionsMultiple Choice Questions

651.

The relation between H and E is :

  • E = H + nRT

  • H = E - nRT

  • H = E + nRT

  • E = H + nRT


652.

In which of the following process , maximum increase in entropy is observed ?

  • Melting of ice

  • Sublimation of naphthalene

  • Condensation of water

  • Dissolution of salt in water


653.

For a reaction to be spontaneous at all temperatures ,

  • G and H till should be negative

  • G and H till should be positive

  • G = S = 0

  • H < G


654.

The heat of formarion of CO2 is - 393 kJ mol-1 .The amount of heat evolved in the formation of 0.156 kg of CO2 .

  • - 1393 kJ

  • + 1165.5 kJ

  • + 1275.9 kJ

  • - 1165.5 kJ


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655.

In an isobaric process, when temperature changes from T1 to T2 ,S is equal to

  • 2.303 Cp log (T2/T1)

  • 2.303 Cp ln (T2/T1)

  • Cp ln (T1/T2)

  • Cv ln (T2/T1)


656.

Heat of neutralisation of an acid with a base is 13. 7 kcal when :

  • both acid and base are weak

  • acid is weak and base is strong

  • both acid and base are strong

  • acid is strong and base is weak


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657.

Hess's law is based on :

  • law of conservation of mass

  • law of conservation of energy

  • enthalpy is a state function

  • none of the above


B.

law of conservation of energy

Hess's law is based upon conservation of energy .


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658.

Assertion : A reaction which is spontaneous and accompanied by decrease of randomness must be exothermic.

Reason : All exothermic reactions are accompanied by decrease of randomness.

  • If both assertion and reason are true and reason is the correct explanation of assertion

  • If both assertion and reason are true but reason is not the correct explanation of assertion

  • If assertion is true but reason is false

  • If both assertion and reason are false


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659.

The enthalpy change (H) for the neutralisation of M HCl by caustic potash in dilute solution at 298 K is :

  • 68 kJ

  • 65 kJ

  • 57.3 kJ

  • 50 kJ


660.

Calculate change in internal energy if H = - 92.2 kJ, P = 40 atm and V = -1L.

  • -42 kJ

  • -88 kJ

  • +88 kJ

  • +42 kJ


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