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Equilibrium

Question
CBSEENCH11008427
Wired Faculty App

The following equilibrium constants are given:

N2 + 3H2  ⇌ 2NH3; K1

N2 +O2  ⇌ 2NO; K2

H2 + 1/2O2  ⇌ H2O' K3

The equilibrium constants for the oxidation of NH3 by oxygen to give NO is:

  • K2K33 /K1

  • KK2K32 /K1

  • K22K3 /K1

  • K1K2 /K3

Solution

A.

K2K33 /K1

The required equation for the oxidation of NH3 by oxygen to give NO is:
4 NH subscript 3 space plus space 5 straight O subscript 2 space rightwards arrow from 800 to the power of straight o space straight C to Pt space left parenthesis gauze right parenthesis of space 4 space NO space plus space 6 straight H subscript 2 straight O For space this space space straight K space equals space fraction numerator left square bracket NO right square bracket to the power of 4 left square bracket straight H subscript 2 straight O right square bracket to the power of 6 over denominator left square bracket NH subscript 3 right square bracket to the power of 4 left square bracket straight O subscript 2 right square bracket to the power of 5 end fraction For space the space equation space left parenthesis straight I right parenthesis space straight K subscript 1 space equals space fraction numerator left square bracket NH subscript 3 right square bracket squared over denominator left square bracket straight N subscript 2 right square bracket left square bracket straight H subscript 2 right square bracket cubed end fraction For space the space equation space left parenthesis II right parenthesis space straight K subscript 2 space equals space fraction numerator left square bracket NO right square bracket squared over denominator left square bracket straight N subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket end fraction For space the space equation space left parenthesis III right parenthesis space straight K subscript 3 space equals space fraction numerator left square bracket straight H subscript 2 straight O right square bracket over denominator left square bracket straight H subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket to the power of 1 divided by 2 end exponent end fraction For space getting space the space straight K space we space must space do straight K subscript 1 squared space equals space fraction numerator left square bracket NH subscript 3 right square bracket to the power of 4 over denominator left square bracket straight N subscript 2 right square bracket squared left square bracket straight H subscript 2 right square bracket to the power of 6 end fraction comma space straight K subscript 2 squared space equals space fraction numerator left square bracket NO right square bracket to the power of 4 over denominator left square bracket straight N subscript 2 right square bracket squared left square bracket straight O subscript 2 right square bracket squared end fraction straight K subscript 3 superscript 6 space equals space fraction numerator left square bracket straight H subscript 2 straight O right square bracket to the power of 6 over denominator left square bracket straight H subscript 2 right square bracket to the power of 6 left square bracket straight O subscript 2 right square bracket to the power of begin display style 6 over 2 end style equals 3 end exponent end fraction straight K space equals space fraction numerator straight K subscript 2 superscript 2 space straight x space straight K subscript 3 superscript 6 over denominator straight K subscript 1 superscript 2 end fraction space substituting space the space value space we space get comma straight K space equals space fraction numerator left square bracket NO right square bracket to the power of 4 left square bracket straight H subscript 2 straight O right square bracket to the power of 6 over denominator left square bracket NH subscript 3 right square bracket to the power of 4 left square bracket straight O subscript 2 right square bracket to the power of 5 end fraction space so space the space value space of space straight K space in space terms space of space straight K subscript 1 comma space straight K subscript 2 space and space straight K subscript 3 space is space space straight K thin space equals space fraction numerator straight K subscript 2 straight K subscript 3 superscript 3 over denominator straight K subscript 1 end fraction

Some More Questions From Equilibrium Chapter

Give an example from daily life in which there is gas solution equilibrium?

A crystal of common salt of a given mass is kept in its aqueous solution. After 24 hours, its mass remains the same. Is the crystal in equilibrium with the solution?

What is the role of forward and backward reactions at equilibrium of a reversible reaction?

What is a forward reaction?

What condition favours a reversible reaction?

What is a backward reaction?

What type of equilibrium is attained in a reversibe chemical change?

What is meant by a system at dynamic equilibrium?

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