Explanation The required reaction is given below. H 2 ( g ) + C l 2 ( g ) ⇌ 2 H C l ( g ) , K c = 5.1 × 10 8 The expression for equilibrium constant for the above reaction can be written as shown below, K c = [ H C l ] 2 [ H 2 ] [ C l 2 ] Given that, the equilibrium concentration of H C l a n d C l 2 is 0.352 m m o l . L − 1 a n d 7.21 m m o l . L − 1 respectively. By rearranging the above expression, the equilibrium concentration of H 2 can be calculated as shown below. K c = [ H C l ] 2 [ H 2 ] [ C l 2 ] [ H 2 ] = [ H C l ] 2 [ C l 2 ] K c [ H 2 ] = ( 0

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Chapter 5, Problem 5I.4E

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Interpretation:

The equilibrium molar concentration of H2 for the given below reaction has to be calculated.

H2(g)+Cl2(g)⇌ 2HCl(g), Kc=5.1×108

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Chapter 5, Problem 5I.3E

Chapter 5, Problem 5I.5E

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Chapter 5 Solutions

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The equilibrium molar concentration of H 2 for the given below reaction has to be calculated. H 2 ( g ) + C l 2 ( g ) ⇌ 2 H C l ( g ) , K c = 5.1 × 10 8

Solution Summary: The author explains the equilibrium molar concentration of H_2 for the given below reaction.

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Interpretation:

The equilibrium molar concentration of H2 for the given below reaction has to be calculated.

H2(g)+Cl2(g)⇌ 2HCl(g), Kc=5.1×108

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Chapter 5 Solutions

The equilibrium molar concentration of H 2 for the given below reaction has to be calculated. H 2 ( g ) + C l 2 ( g ) ⇌ 2 H C l ( g ) , K c = 5.1 × 10 8

Solution Summary: The author explains the equilibrium molar concentration of H_2 for the given below reaction.

Videos

Interpretation: The equilibrium molar concentration of H2 for the given below reaction has to be calculated. H2(g)+Cl2(g)⇌ 2HCl(g), Kc=5.1×108

Want to see the full answer?

Chapter 5 Solutions

Interpretation:

The equilibrium molar concentration of H2 for the given below reaction has to be calculated.

H2(g)+Cl2(g)⇌ 2HCl(g), Kc=5.1×108