Manipulating the Equilibrium Constant (College Board AP® Chemistry)

2BrCl (g) ⇌ Br₂ (g) + Cl₂ (g)           K_{c(500 K)} = 32

Based on the information above, determine the value of K_c for the following reactions at 500 K.

1. Br₂ (g) + Cl₂ (g) ⇌ 2BrCl (g)

2. 6BrCl (g) ⇌ 3Br₂ (g) + 3Cl₂ (g)

3. ½Br₂ (g) + ½Cl₂ (g) ⇌ BrCl (g)

Answers:

Answer 1:

• The new equation is the original equation in reverse

• The new K_c will be the inverse of the original

Answer 2:

• The stoichiometric coefficients in the new equation are three times larger than those in the original

• The new K_c will be equal to the original K_c raised to the power of 3

Answer 3:

• The new equation is the original equation in reverse but the stoichiometric coefficients in the new equation are one-half those in the original

• The new K_c will be the inverse of the original K_c raise to the power of ½

 Based on the information above, calculate the equilibrium constant, K_P, at 1300 K for the reaction below.

C₆H₁₂ (g) ⇌ 3C₂H₂ (g) + 3H₂ (g)

Answer:

Step 1: Manipulating the first equation

• The desired equation contains three moles of C₂H₂ and three moles of H₂ as products

• To achieve this, the stoichiometric coefficients in the first equation must be multiplied by three

3C₂H₄ (g) ⇌ 3C₂H₂ (g) + 3H₂ (g)

• The new K_c for this equation will be equal to the original K_c raised to the power of 3

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Step 2: Manipulating the second equation

• The desired equation contains one mole of C₆H₁₂ as a reactant

• To achieve this, the second equation must be reversed

C₆H₁₂ (g) ⇌ 3C₂H₄ (g)

• The new K_c will be the inverse of the original

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Step 3: Combining the equations

• Combining the new equations gives the desired equation

3C₂H₄ (g) + C₆H₁₂ (g) ⇌ 3C₂H₂ (g) + 3H₂ (g) + 3C₂H₄ (g)

C₆H₁₂ (g) ⇌ 3C₂H₂ (g) + 3H₂ (g)

• The K_c of the desired equation is the product of the K_c’s of the summed equations

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Be careful not to confuse the rules of manipulating equilibrium constants with those of changes in enthalpy. When the stoichiometric coefficients of a reaction are multiplied by a factor, c, the change in enthalpy of the reaction is multiplied by c but the equilibrium constant is raised to the power of c. This is shown in the table below.