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Equilibrium

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Chemistry Part I

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Chemistry

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Class 10 Class 12

For a given exothermic reaction Kp and Kp' are the equilibrium constant at temperatures T1 and T2 respectively. Assuming that heat of reaction si constant in temperature range between T1 and T2 it is readily observed that

  • Kp> Kp'

  • Kp< Kp'

  • Kp = Kp'

  • Kp =1/ Kp'


A.

Kp> Kp'

The equilibrium constant at two different temperatures for a thermodynamic process is given by

log straight K subscript 2 over straight K subscript 1 space equals space fraction numerator increment straight H over denominator 2.303 space straight R end fraction space open square brackets 1 over straight T subscript 1 minus 1 over straight T subscript 2 close square brackets
here space straight K subscript 1 space amd space straight K subscript 2 space are space replaced space by space straight K subscript straight p space and space straight K subscript straight p superscript apostrophe.
Therefore comma space log space fraction numerator straight K subscript straight p superscript apostrophe over denominator straight K subscript straight p end fraction space equals space fraction numerator increment straight H to the power of degree over denominator 2.303 straight R end fraction open square brackets 1 over straight T subscript 1 minus 1 over straight T subscript 2 close square brackets
For space exothermic space reaction comma space straight T subscript 2 space greater than straight T subscript 1 space and space increment straight H space equals space minus ve space
straight K subscript straight p greater than space straight K subscript straight p superscript apostrophe

The equilibrium constant at two different temperatures for a thermodynamic process is given by

log straight K subscript 2 over straight K subscript 1 space equals space fraction numerator increment straight H over denominator 2.303 space straight R end fraction space open square brackets 1 over straight T subscript 1 minus 1 over straight T subscript 2 close square brackets
here space straight K subscript 1 space amd space straight K subscript 2 space are space replaced space by space straight K subscript straight p space and space straight K subscript straight p superscript apostrophe.
Therefore comma space log space fraction numerator straight K subscript straight p superscript apostrophe over denominator straight K subscript straight p end fraction space equals space fraction numerator increment straight H to the power of degree over denominator 2.303 straight R end fraction open square brackets 1 over straight T subscript 1 minus 1 over straight T subscript 2 close square brackets
For space exothermic space reaction comma space straight T subscript 2 space greater than straight T subscript 1 space and space increment straight H space equals space minus ve space
straight K subscript straight p greater than space straight K subscript straight p superscript apostrophe

1365 Views . 11 Shares

Given the reaction between two gases represented by A2 and B2 to give the compound AB (g).
A2(g) +B2 (g)  ⇌ 2AB (g)
At equilibrium the concentration
of A2 = 3.0 x 10-3 M
of B2 = 4.2 x 10-3 M
of AB = 2.8 x 10-3 M
If the reaction takes place in a sealed vessel at 527oC, then the value of Kc will be

  • 2.0

  • 1.9

  • 0.62

  • 4.5


C.

0.62

A2 (g) +B2 (g)  ⇌ 2AB (g)
The equilibrium constant is given by

straight K subscript straight c space equals space fraction numerator left square bracket AB right square bracket squared over denominator left square bracket straight A subscript 2 right square bracket left square bracket straight B subscript 2 right square bracket end fraction space equals
straight K subscript straight c space equals space fraction numerator left parenthesis 2.8 space straight x space 10 to the power of negative 3 end exponent right parenthesis squared over denominator left parenthesis 3.0 space straight x space space 10 to the power of negative 3 end exponent right parenthesis left parenthesis 4.2 space straight x space 10 to the power of negative 3 end exponent right parenthesis end fraction
straight K subscript straight c space equals space fraction numerator 7.84 over denominator 12.6 end fraction
space equals space 0.62

A2 (g) +B2 (g)  ⇌ 2AB (g)
The equilibrium constant is given by

straight K subscript straight c space equals space fraction numerator left square bracket AB right square bracket squared over denominator left square bracket straight A subscript 2 right square bracket left square bracket straight B subscript 2 right square bracket end fraction space equals
straight K subscript straight c space equals space fraction numerator left parenthesis 2.8 space straight x space 10 to the power of negative 3 end exponent right parenthesis squared over denominator left parenthesis 3.0 space straight x space space 10 to the power of negative 3 end exponent right parenthesis left parenthesis 4.2 space straight x space 10 to the power of negative 3 end exponent right parenthesis end fraction
straight K subscript straight c space equals space fraction numerator 7.84 over denominator 12.6 end fraction
space equals space 0.62

1200 Views . 11 Shares

If the value of an equilibrium constant ofr particular reaction is 1.6 x1012 then at equilibrium the system will contains

  • all reactants

  • mostly reactants

  • mostly products

  • similar amounts of reactants and products


C.

mostly products

For a reaction,

stack straight A space with Reactant below leftwards harpoon over rightwards harpoon space straight B with product below

straight K space equals space fraction numerator left square bracket straight B right square bracket subscript eq over denominator left square bracket straight A right square bracket space subscript eq end fraction space equals space 1.6 space straight x space 10 to the power of 12 space equals space fraction numerator left square bracket straight B right square bracket subscript eq over denominator left square bracket straight A right square bracket subscript eq end fraction

therefore space left square bracket straight B right square bracket subscript eq space greater than greater than space left square bracket straight A right square bracket subscript eq
So, mostly the product will be present in the equilibrium mixture.

For a reaction,

stack straight A space with Reactant below leftwards harpoon over rightwards harpoon space straight B with product below

straight K space equals space fraction numerator left square bracket straight B right square bracket subscript eq over denominator left square bracket straight A right square bracket space subscript eq end fraction space equals space 1.6 space straight x space 10 to the power of 12 space equals space fraction numerator left square bracket straight B right square bracket subscript eq over denominator left square bracket straight A right square bracket subscript eq end fraction

therefore space left square bracket straight B right square bracket subscript eq space greater than greater than space left square bracket straight A right square bracket subscript eq
So, mostly the product will be present in the equilibrium mixture.

1313 Views . 8 Shares

For a spontaneous reaction the ∆G, equilibrium constant (K) and Eocell will be respectively 

  • -ve, >1, +ve

  • +ve, >1, -ve

  • -ve, <1, -ve

  • -ve, >1, -ve


A.

-ve, >1, +ve

169 Views . 10 Shares

The exothermic formation of ClF3 is represented by the equation
Cl subscript 2 left parenthesis straight g right parenthesis space plus space 3 straight F subscript 2 space left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon with space space space space space space space on top space 2 ClF subscript 3 space left parenthesis straight g right parenthesis space semicolon space increment rH space equals negative 329 space kJ
Which of the following will increase the quantity of ClF3 in an equilibrium mixture of Cl2, F2 and ClF3?

  • Increasing the temperature

  • Removing Cl2

  • Increasing the volume of the container

  • Adding F2


D.

Adding F2

straight M subscript 3 straight V subscript 3 space equals space straight M subscript 1 straight V subscript 2 space plus space straight M subscript 2 straight V subscript 2
straight M space equals space fraction numerator 480 space left parenthesis 1.5 right parenthesis space plus space 520 space left parenthesis 1.2 right parenthesis over denominator 1000 end fraction space equals space 1.344 space straight M
straight M subscript 3 straight V subscript 3 space equals space straight M subscript 1 straight V subscript 2 space plus space straight M subscript 2 straight V subscript 2
straight M space equals space fraction numerator 480 space left parenthesis 1.5 right parenthesis space plus space 520 space left parenthesis 1.2 right parenthesis over denominator 1000 end fraction space equals space 1.344 space straight M
425 Views . 4 Shares

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