Question

bromine and water react to form hydrogen bromide and oxygen, like this:
2br₂(g) + 2h₂o(g) → 4hbr(g) + o₂(g)
also, a chemist finds that at a certain temperature the equilibrium mixture of bromine, water, hydrogen bromide, and oxygen has the following composition:

calculate the value of the equilibrium constant kₚ for this reaction. round your answer to 2 significant digits.
kₚ = □

Explanation:

Step1: Write the formula for \(K_p\)

For the reaction \(2Br_2(g)+2H_2O(g)\to4HBr(g) + O_2(g)\), the formula for \(K_p\) is \(K_p=\frac{P_{HBr}^4\times P_{O_2}}{P_{Br_2}^2\times P_{H_2O}^2}\)

Step2: Substitute the values

Substitute \(P_{Br_2} = 63.4\ atm\), \(P_{H_2O}=60.2\ atm\), \(P_{HBr}=21.8\ atm\), and \(P_{O_2}=84.8\ atm\) into the formula:

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First, calculate \((21.8)^4=21.8\times21.8\times21.8\times21.8 = 21.8^2\times21.8^2=(475.24)\times(475.24)=225852.0576\)
\((63.4)^2 = 63.4\times63.4 = 4019.56\)
\((60.2)^2=60.2\times60.2 = 3624.04\)

Then \(K_p=\frac{225852.0576\times84.8}{4019.56\times3624.04}\)

\(225852.0576\times84.8 = 225852.0576\times(80 + 4.8)=225852.0576\times80+225852.0576\times4.8=18068164.608+1084089.87648 = 19152254.48448\)

\(4019.56\times3624.04=(4000 + 19.56)\times3624.04=4000\times3624.04+19.56\times3624.04=14496160+69986.2224 = 14566146.2224\)

\(K_p=\frac{19152254.48448}{14566146.2224}\approx1.3\)

Answer:

\(1.3\)