JEE chemistry

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Question
CBSEENCH11008127

The equilibrium constants KP1 and KP2 for the reactions X ⇌2Y and Z ⇌ P + Q, respectively are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressure at these equilibria is

  • 1:1

  • 1:36

  • 1:9

  • 1:3

Solution

A.

1:1

straight X space space rightwards harpoon over leftwards harpoon with space space space space space space space space space space space space space on top space 2 straight Y
stack 1 space with left parenthesis 1 minus straight x right parenthesis below space space space space space space space space space space space 0 with space 2 straight x below
straight k subscript straight p subscript 1 end subscript space equals space fraction numerator left parenthesis 2 straight x right parenthesis squared over denominator left parenthesis 1 minus straight x right parenthesis end fraction open parentheses fraction numerator straight P subscript 1 over denominator 1 plus straight x end fraction close parentheses
straight Z space rightwards harpoon over leftwards harpoon with space on top space space space space space space straight P plus space straight Q
1 with left parenthesis 1 minus straight x right parenthesis below space space space space space space space space space space 0 with straight x below space space space space space space space 0 with straight x below
straight k subscript straight p subscript 2 space equals space end subscript fraction numerator left parenthesis straight x right parenthesis squared over denominator left parenthesis 1 minus straight x right parenthesis end fraction open parentheses fraction numerator straight P subscript 1 over denominator 1 plus straight x end fraction close parentheses to the power of 1
fraction numerator 4 space straight x space straight P subscript 1 over denominator straight P subscript 2 end fraction space equals space 1 over 9 space
rightwards double arrow space straight P subscript 1 over straight P subscript 2 space equals space 1 over 36

Sponsor Area

Question
CBSEENCH11008128

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-

  • +120  kJ mol-

Solution

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

Question
CBSEENCH11008129

For the following three reactions a, b and c, equilibrium constants are given

a. CO (g) + H2O (g) ⇌ CO2 (g)+ H2 (g); K1
b. CH4(g) + H2O (g) ⇌ CO (g) + 3H2 (g); K2
c. CH4(g) + 2H2O (g) ⇌ CO2 (g) + 4H2 (g); K2
Which of the following relations is correct?

  • K1√K2 = K3

  • K2K3 = K1

  • K3 = K1K2

  • K3.K32 = K12

Solution

C.

K3 = K1K2

Equation (c) = equation (a) + equation (b) Thus K3 = K1.K2

Question
CBSEENCH11008131

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

  • 1000 K

Solution

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

Question
CBSEENCH11008132

Four species are listed below
i. HCO3
ii. H3O+
iii. HSO4
iv. HSO3F
Which one of the following is the correct sequence of their acid strength?

  •  iv < ii < iii < i

  • ii < iii < i < iv

  • i < iii < ii < iv

  • iii < i < iv < ii 

Solution

C.

i < iii < ii < iv