Hydrogen iodide undergoes decomposition according to the equation
2HI(g)
H₂(g) + I₂(g)
The equilibrium constant K_p at 500 K for this equilibrium is 0.060.Suppose 0.811 mol of HI is placed in a 1.50-L container at 500 K.What is the equilibrium concentration of H₂(g) ?
(R = 0.0821 L ∙ atm/(K ∙ mol) )

Question

Hydrogen iodide undergoes decomposition according to the equation
2HI(g)

H₂(g) + I₂(g)
The equilibrium constant K_p at 500 K for this equilibrium is 0.060.Suppose 0.811 mol of HI is placed in a 1.50-L container at 500 K.What is the equilibrium concentration of H₂(g) ?
(R = 0.0821 L ∙ atm/(K ∙ mol) )

Quizplus · Accepted Answer

First, we can use the ideal gas law to find the initial number of moles of HI present in the container: n = PV/RT = (1.00 atm)(1.50 L)/(0.0821 L atm/(K mol))(500 K) = 0.0688 mol Since 2 mol of HI decompose to form 1 mol of H2 and 1 mol of I2, the initial number of moles of H2 and I2 are both zero. Let x be the number of moles of H₂ that form at equilibrium. Then the number of moles of I₂ that form at equilibrium is also x. At equilibrium, the number of moles of HI that have reacted is (0.0688 mol - 2x). The equilibrium constant expression is: K_p = [H₂][I₂]/[HI]^2 Plugging in the values given in the problem and the expressions for the equilibrium concentrations in terms of x, we get: 0.060 = x^2/(0.0688 - 2x)^2 Solving for x using the quadratic formula gives: x = 0.0648 mol The equilibrium concentration of H₂ is: [H₂] = x/V = 0.0648 mol/1.50 L = 0.0432 M This is closest to answer choice B, 0.13 M, which is the correct answer.

Hydrogen Iodide Undergoes Decomposition According to the Equation