Question

bromine is one of the few elements that is a liquid at room temperature. it vaporizes easily and has a beautiful red color in both the liquid and gaseous states.
bromine reacts readily with hydrogen gas to make hydrogen bromide.
one way to represent this equilibrium is:
hbr(g) 1/2 h₂(g) + 1/2 br₂(g)
we could also write this reaction three other ways, listed below. the equilibrium constants for all of the reactions are related. write the equilibrium constant for each new reaction in terms of k, the equilibrium constant for the reaction above.

1. 2 hbr(g) h₂(g) + br₂(g)

k₁ =

1. h₂(g) + br₂(g) 2hbr(g)

k₂ =
3)1/2 h₂(g) + 1/2 br₂(g) hbr(g)
k₃ =
drag and drop your selection from the following list to complete the answer:

Explanation:

Step1: Recall the relationship between equilibrium constants and stoichiometry

If a reaction is reversed, the new equilibrium constant \(K'\) is the reciprocal of the original equilibrium constant \(K\). If a reaction is multiplied by a factor \(n\), the new equilibrium constant \(K'\) is \(K^{n}\).

Step2: Analyze reaction 2

The original reaction is \(H_{2}(g)+Br_{2}(g)
ightleftharpoons 2HBr(g)\) with equilibrium constant \(K\). Reaction 2 is \(2HBr(g)
ightleftharpoons H_{2}(g)+Br_{2}(g)\), which is the reverse of the original reaction. So \(K_{2}=\frac{1}{K}\)

Step3: Analyze reaction 3

Reaction 3 is \(\frac{1}{2}H_{2}(g)+\frac{1}{2}Br_{2}(g)
ightleftharpoons HBr(g)\). The original reaction \(H_{2}(g)+Br_{2}(g)
ightleftharpoons 2HBr(g)\) is multiplied by \(\frac{1}{2}\). Using the rule \(K' = K^{n}\) (where \(n=\frac{1}{2}\)), if the original equilibrium constant is \(K\), for the reaction \(aA + bB
ightleftharpoons cC\) with \(K=\frac{[C]^{c}}{[A]^{a}[B]^{b}}\), for the reaction \(\frac{a}{2}A+\frac{b}{2}B
ightleftharpoons\frac{c}{2}C\), \(K'=\sqrt{K}\)

Answer:

\(K_{1}=K\), \(K_{2}=\frac{1}{K}\), \(K_{3}=\sqrt{K}\)