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The equilibrium constant (Kc) for the reaction 2HCl (g) <-->H2(g) Cl2(g) is 4×10^-34 at 25 degree C .what is the equilibrium constant for the reaction :- 1/2H2 (g) Cl2(g) <---> HCl(g).?
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The equilibrium constant (Kc) for the reaction 2HCl (g) H2(g) Cl2(g)...
Understanding Equilibrium Constants
The equilibrium constant (Kc) represents the ratio of the concentration of products to reactants at equilibrium. For the reaction:
2HCl (g) ⇌ H2(g) + Cl2(g),
the given Kc is 4 × 10^-34.
Reversing the Reaction
When we reverse a reaction, the equilibrium constant is inverted. Thus, for the reaction:
H2(g) + Cl2(g) ⇌ 2HCl(g),
the equilibrium constant (Kc') is:
Kc' = 1 / (4 × 10^-34) = 2.5 × 10^33.
Changing the Reaction Stoichiometry
Now, we need to find the equilibrium constant for the reaction:
1/2H2(g) + Cl2(g) ⇌ HCl(g).
To adjust the Kc for this new reaction, we recognize that it is half of the reversed reaction. According to the equilibrium constant rules, when the stoichiometric coefficients are multiplied by a factor, the equilibrium constant is raised to the power of that factor.
Since we have halved the coefficients (1/2), we take the square root of Kc':
Kc'' = (Kc')^(1/2) = (2.5 × 10^33)^(1/2) = 5 × 10^16.
Final Equilibrium Constant
Therefore, the equilibrium constant for the reaction:
1/2H2(g) + Cl2(g) ⇌ HCl(g)
is:
Kc = 5 × 10^16.
This indicates a strong tendency for the formation of HCl at equilibrium.
The equilibrium constant (Kc) represents the ratio of the concentration of products to reactants at equilibrium. For the reaction:
2HCl (g) ⇌ H2(g) + Cl2(g),
the given Kc is 4 × 10^-34.
Reversing the Reaction
When we reverse a reaction, the equilibrium constant is inverted. Thus, for the reaction:
H2(g) + Cl2(g) ⇌ 2HCl(g),
the equilibrium constant (Kc') is:
Kc' = 1 / (4 × 10^-34) = 2.5 × 10^33.
Changing the Reaction Stoichiometry
Now, we need to find the equilibrium constant for the reaction:
1/2H2(g) + Cl2(g) ⇌ HCl(g).
To adjust the Kc for this new reaction, we recognize that it is half of the reversed reaction. According to the equilibrium constant rules, when the stoichiometric coefficients are multiplied by a factor, the equilibrium constant is raised to the power of that factor.
Since we have halved the coefficients (1/2), we take the square root of Kc':
Kc'' = (Kc')^(1/2) = (2.5 × 10^33)^(1/2) = 5 × 10^16.
Final Equilibrium Constant
Therefore, the equilibrium constant for the reaction:
1/2H2(g) + Cl2(g) ⇌ HCl(g)
is:
Kc = 5 × 10^16.
This indicates a strong tendency for the formation of HCl at equilibrium.
Community Answer
The equilibrium constant (Kc) for the reaction 2HCl (g) H2(g) Cl2(g)...
K2 = 1/√K1. K2 = 1 / √4*10-34. K2 = 5*1016
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The equilibrium constant (Kc) for the reaction 2HCl (g) H2(g) Cl2(g) is 4×10^-34 at 25 degree C .what is the equilibrium constant for the reaction :- 1/2H2 (g) Cl2(g) HCl(g).?
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The equilibrium constant (Kc) for the reaction 2HCl (g) H2(g) Cl2(g) is 4×10^-34 at 25 degree C .what is the equilibrium constant for the reaction :- 1/2H2 (g) Cl2(g) HCl(g).? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The equilibrium constant (Kc) for the reaction 2HCl (g) H2(g) Cl2(g) is 4×10^-34 at 25 degree C .what is the equilibrium constant for the reaction :- 1/2H2 (g) Cl2(g) HCl(g).? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equilibrium constant (Kc) for the reaction 2HCl (g) H2(g) Cl2(g) is 4×10^-34 at 25 degree C .what is the equilibrium constant for the reaction :- 1/2H2 (g) Cl2(g) HCl(g).?.
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