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251.
For the reaction $2 Cl left parenthesis straight g right parenthesis space rightwards arrow space space Cl subscript 2 left parenthesis straight g right parenthesis comma$ what are the signs of $increment straight H space and space increment straight S.$ ?

Since the given reaction

represents the formation of bonds, therefore energy is released i.e. ∆H is –ve, further 2 moles of atoms have greater randomness than 1 mole of molecules. Hence randomness decreases, i.e. ∆S is < 0 i.e. negative.

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Long Answer Type

252.

Prove that in a reversible process:
∆_(system) ⁺ ∆S_{(surroundings)} = 0
Or
Prove that there is no net change in entropy in a reversible process.

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253.

Prove that in an irreversible process:
∆S_(system) + ∆S_{(surroundings)}> 0

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254.

What do you understand by:
(i) The entropy of fusion?
(ii) The entropy of vapourisation ?

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Short Answer Type

255.

You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below:
Liquid $straight b. straight p. space left parenthesis degree straight C right parenthesis space space space space space space space space space increment straight H subscript vap superscript 0 space end superscript space left parenthesis kJ space mol to the power of negative 1 end exponent right parenthesis$
$space left parenthesis straight a right parenthesis space Ethanol with left parenthesis straight C subscript 2 straight H subscript 5 OH right parenthesis below space space space space space space space space space space space space space space space space space space 78.4 space space space space space space space space space 42.4 space left parenthesis straight b right parenthesis space Toluene with left parenthesis straight C subscript 2 straight H subscript 5 CH subscript 3 right parenthesis below space space space space space space space space space space space space space space space space space space 110.6 space space space space space space space 35.2 space space$

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256.

Calculate the entropy change of n-hexane when 1 mole of it evaporates at 341.7 K(∆H_vap = 29.0 kJ mol^–1).

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257. When one mole of a solid ‘S’ (molecular weight 46) at its melting point is changed into the liquid at the same temperature, an entropy change of 26.4 JK^–1takes place. Calculate the melting point of the solid if its enthalpy of fusion is 0 .1 kJg^–1.

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