Consider the equilibrium reaction between the gas phases of hydrogen and bromine: H₂ (g) + Br₂ (g) $\rightarrow$ 2 HBr (g)
If the partial pressures of hydrogen and of bromine are 0.016 atm each,what is the partial pressure of HBr if K = 2.2 × 10³?

Question

Consider the equilibrium reaction between the gas phases of hydrogen and bromine: H₂ (g) + Br₂ (g)

\rightarrow

2 HBr (g)
If the partial pressures of hydrogen and of bromine are 0.016 atm each,what is the partial pressure of HBr if K = 2.2 × 10³?

Quizplus · Accepted Answer

First, we need to calculate the reaction quotient Qp using the partial pressures given: Qp = (PH2)(PBr2) / (PHBr)2 Qp = (0.016 atm)(0.016 atm) / (x)^2 Qp = 0.000256 atm2 / (x)^2 Next, we need to compare Qp to the equilibrium constant K: If Qp < K, the reaction will shift to the right to reach equilibrium, producing more HBr. If Qp > K, the reaction will shift to the left to reach equilibrium, producing more H2 and Br2. If Qp = K, the reaction is already at equilibrium. So, plugging in Qp and K: 0.000256 atm2 / (x)^2 < 2.2 x 10^13 Simplifying, we get: x^2 > 1.164 x 10^-10 x > 3.41 x 10^-6 Therefore, the partial pressure of HBr must be greater than 3.41 x 10^-6 atm. The only answer choice that satisfies this is A) 0.75 atm.

Consider the Equilibrium Reaction Between the Gas Phases of Hydrogen →\rightarrow→

Consider the Equilibrium Reaction Between the Gas Phases of Hydrogen $\rightarrow$