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 Multiple Choice QuestionsMultiple Choice Questions

21.

If an endothermic reaction occurs spontaneously at constant temperature (T) and pressure (p), then which of the following is true?

  • G > 0

  • H < 0

  • S > 0

  • S < 0


C.

S > 0

For endothermic reaction H is positive ie, H >0. For a spontaneous process G is always Negative and G = H - TS thus, to maintain the value of G negative, S must be postive ie, S>0.


22.

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V2 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

  • fraction numerator straight T subscript 1 straight T subscript 2 space left parenthesis straight P subscript 1 straight V subscript 1 space plus space straight P subscript 2 straight V subscript 2 right parenthesis over denominator straight P subscript 1 straight V subscript 1 straight T subscript 2 space plus space straight P subscript 2 straight V subscript 2 straight T subscript 1 end fraction
  • fraction numerator left parenthesis straight P subscript 1 straight V subscript 1 space straight T subscript 1 plus space straight P subscript 2 straight V subscript 2 straight T subscript 2 right parenthesis over denominator straight P subscript 1 straight V subscript 1 space plus space straight P subscript 2 straight V subscript 2 end fraction
  • fraction numerator left parenthesis straight P subscript 1 straight V subscript 1 space straight T subscript 2 plus space straight P subscript 2 straight V subscript 2 straight T subscript 1 right parenthesis over denominator straight P subscript 1 straight V subscript 1 space plus space straight P subscript 2 straight V subscript 2 end fraction
  • fraction numerator left parenthesis straight P subscript 1 straight V subscript 1 space straight T subscript 2 plus space straight P subscript 2 straight V subscript 2 straight T subscript 1 right parenthesis over denominator straight P subscript 1 straight V subscript 1 space plus space straight P subscript 2 straight V subscript 2 end fraction

A.

fraction numerator straight T subscript 1 straight T subscript 2 space left parenthesis straight P subscript 1 straight V subscript 1 space plus space straight P subscript 2 straight V subscript 2 right parenthesis over denominator straight P subscript 1 straight V subscript 1 straight T subscript 2 space plus space straight P subscript 2 straight V subscript 2 straight T subscript 1 end fraction
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23.

Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below

1 half Cl subscript 2 space left parenthesis straight g right parenthesis space rightwards arrow with 1 half space increment subscript diss straight H to the power of minus on top space Cl space left parenthesis straight g right parenthesis space rightwards arrow with increment subscript eg straight H to the power of minus on top space Cl to the power of minus space left parenthesis straight g right parenthesis space rightwards arrow with increment subscript hyd straight H to the power of minus on top Cl to the power of minus space left parenthesis aq right parenthesis
The energy involved in conversion of 1 half Cl subscript 2 space left parenthesis straight g right parenthesis space to space Cl to the power of minus space left parenthesis straight g right parenthesis

left parenthesis using space the space data comma space increment subscript diss straight H subscript cl subscript 2 end subscript superscript minus space equals space 240 space kJ space mol to the power of minus comma space increment subscript eg space straight H subscript cl superscript minus space equals space minus space 349 space kJ space mol to the power of minus comma space
increment subscript eg space straight H subscript cl superscript minus space equals space minus 381 space kJ space mol to the power of minus right parenthesis

  • 152 kJ mol-

  • -610 kJ mol-

  • -850  kJ mol-

  • -850  kJ mol-


B.

-610 kJ mol-

1 half space Cl subscript 2 space left parenthesis straight g right parenthesis space rightwards arrow with space on top space Cl subscript aq superscript minus
space increment space straight H space equals space 1 half space increment straight H subscript diss space of space Cl subscript 2 space plus space increment subscript eq space Cl space plus space increment subscript hyd Cl to the power of minus
equals space plus space 240 over 2 space minus 349 minus 381
space equals space minus space 610 space kJ space mol to the power of negative 1 end exponent
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24.

Condition for spontaneity in an isothermal process is

  • A + W < 0

  • G + U <0

  • A + U>0

  • G -U<0


A.

A + W < 0

The condition for spontaneity in an isothermal process is that G should be negative.

 G =A +P·VG =A +W     ( W= P·V)For spontaneity in an isothermal process: A + W <0


25.

2C(s) + 2O2(g)  2CO2(g);              H= -787 kJ ....(i)H2(g) + 12O2(g)  H2O (l);              H = -286 kJ ...(ii)C2H2(g) + 212O2(g)  2CO2(g) + H2O (l) ;  H= -1310 kJ ...(iii)

The heat of formation of acetylene is :

  • -1802 kJ

  • +1802 kJ

  • +237 kJ

  • -800 kJ


C.

+237 kJ

2C(s) + 2O2(g)  2CO2(g);              H= -787 kJ ....(i)H2(g) + 12O2(g)  H2O (l);              H = -286 kJ ...(ii)C2H2(g) + 212O2(g)  2CO2(g) + H2O (l) ;  H= -1310 kJ ...(iii)Add the Eqs. (i) and (ii) and subtract Eq. (iii) from them. 2C + H2 C2H2 - 787 -286+13102C + H2  C2H2- 1073 + 13102C + H2 C2H2 + 237Hence, the heat of formation of acetylene is + 237 kJ.


26.

Which of the following is a path function?

  • Internal energy

  • Enthalpy

  • Work

  • Entropy


C.

Work

Internal energy, enthalpy and entropy are state functions but work and heat are path function.


27.

Hess's law is based on

  • law of conservation of mass

  • law of conservation of energy

  • first law of thermodynamics

  • None of the above


B.

law of conservation of energy

Hess's law is based upon law of conservation of energy ie, first law of thermodynamics.


28.

Enthalpy is equal to :

  • T2(G/T)TP

  • -T2(G/T)TP

  • T2(G/T)TV

  • -T2(G/T)TV


B.

-T2(G/T)TP

The Gibbs-Helmholtz equation is as :   G= H + T GTPDividing above equation by T2GT2= HT2+ 1TGTThis on rearrangement becomes (G/T)TP=HT2H= -T2(G/T)TPWhere H= enthalpy


29.

For the reaction,

2H2(g) + O2(g)  2H2O(g), H°= - 573.2 kJ

The heat of decomposition of water per mole is

  • 286.6 kJ

  • 573.2 kJ

  • -28.66 kJ

  • Zero


A.

286.6 kJ

Heat of decomposition of water is

H2O(g)  H2(g)+ 12O2(g),H=+ 573.22= 286.6 kJ/mol


30.

Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK−1
mol −1, respectively. For the reaction,1/2X2 + 3/2Y2, ΔH = -30 kJ,to be at equilibrium, the temperature will be

  • 1250 K

  • 500 K 

  • 750 K

  • 750 K


C.

750 K

1 half space straight X subscript 2 space plus space 3 over 2 straight Y subscript 2 space rightwards arrow space XY subscript 3
space increment straight S subscript reaction space equals space minus space 50 minus open parentheses 3 over 2 space straight x space 40 space plus space 1 half space straight x space 60 close parentheses space equals space minus space 40 space straight J space mol to the power of negative 1 end exponent
increment straight G space space equals space increment straight H space minus space straight T increment straight S
at space equilibrium space increment straight G space equals space 0
increment straight H space equals space straight T increment straight S
30 space straight x space 10 cubed space equals space straight T space straight x space 40
rightwards double arrow space straight T space equals space 750 space straight K
392 Views