At 298 K,the value of K_c for the reaction H₂(g) + Br₂(g) 2HBr(g) is 2.0 × 10¹⁹.What is K_c for HBr(g)
1/2H₂(g) + 1/2Br₂(g) ?

Question

At 298 K,the value of K_c for the reaction H₂(g)+ Br₂(g) 2HBr(g)is 2.0 × 10¹⁹.What is K_c for HBr(g) 1/2H₂(g)+ 1/2Br₂(g)?

Quizplus · Accepted Answer

The equilibrium constant, Kc, for a reaction is the product of the concentrations of products raised to their stoichiometric coefficients in the balanced chemical equation, divided by the product of the concentrations of reactants raised to their stoichiometric coefficients:
Kc = [HBr]^2 / [HSBr2][Br2]

We can use the equilibrium constant, Kc, for the reaction given in the problem to find the equilibrium constant for the reaction we are interested in, which is the reverse reaction:
Kc = 2.0 × 10^19 = ([HBr]^2 / [HSBr2][Br2])

We are given the equation for the reverse reaction: 
1/2HSBr2 + 1/2Br2 ⇌ HBr

We can write the equilibrium constant, Kc', for this reaction using the law of mass action:
Kc' = [HBr] / ([HSBr2]^(1/2) [Br2]^(1/2))

We can substitute the equilibrium concentrations of HSBr2 and Br2 in terms of the concentration of HBr using the stoichiometry of the two reactions:
[HSBr2] = (1/2)[HS2][Br2]
[Br2] = (1/2)[HS2][Br2]

Substituting these into the equation for Kc', we get:
Kc' = [HBr] / ([(1/2)[HS2][Br2]]^(1/2) [(1/2)[HS2][Br2]]^(1/2))

Simplifying, we get:
Kc' = [HBr] / [(1/4)[HS2][Br2]]

We can substitute the equilibrium constant, Kc, into this equation and solve for Kc':
2.0 × 10^19 = ([HBr]^2 / [(1/2)[HS2][Br2]]^[1/2] [(1/2)[HS2][Br2]]^[1/2])
[HBr]^2 = 2.0 × 10^19 [(1/4)[HS2][Br2]]
[HBr] = (2.2 × 10^-10) [(HS2][Br2])^[1/2]

Substituting this expression for [HBr] into the equation for Kc', we get:
Kc' = [(2.2 × 10^-10) [(HS2][Br2])^[1/2]] / [(1/4)[HS2][Br2]]
Kc' = 8.8 × 10^-11

This is equivalent to 2.2 × 10^-10 / (1/2) / (1/2), which matches answer choice E.

At 298 K,the Value of Kc for the Reaction H2(g)

At 298 K,the Value of K_c for the Reaction H₂(g)