Consider the equilibrium reaction between the gas phases of hydrogen and iodine: H₂ (g) + I₂ (g) $\rightarrow$ 2 HI (g)
If the partial pressures of hydrogen and of iodine are 0.42 atm each,what is the partial pressure of HI if K = 7.1 × 10²?

Question

Consider the equilibrium reaction between the gas phases of hydrogen and iodine: H₂ (g) + I₂ (g)

\rightarrow

2 HI (g)
If the partial pressures of hydrogen and of iodine are 0.42 atm each,what is the partial pressure of HI if K = 7.1 × 10²?

Quizplus · Accepted Answer

The equilibrium constant expression for this reaction is:K = [HI]^2 / [H2] [I2]Substituting the given partial pressures and the equilibrium constant into this expression, we get:7.1 × 10^2 = [HI]^2 / (0.42 atm)^2Solving for [HI], we get:[HI] = 11 atmTherefore, the partial pressure of HI is 11 atm.

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

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