Hydrogen iodide decomposes according to the equation: 2HI(g) ⇌ H₂(g) + I₂(g) , K_c = 0.0156 at 400ºC
a 0.660 mol sample of HI was injected into a 2.00 L reaction vessel held at 400ºC.Calculate the concentration of HI at equilibrium.

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is:K_c = [H₂][I₂]/[HI]²At equilibrium, the concentrations of the reactants and products will be such that the value of K_c is equal to the value given in the problem.We can use the ICE table to determine the equilibrium concentrations of the reactants and products:```Initial: [HI] = 0.660 M, [H₂] = 0 M, [I₂] = 0 MChange: -2x, +x, +xEquilibrium: [HI] = 0.660 - 2x, [H₂] = x, [I₂] = x```Substituting the equilibrium concentrations into the equilibrium constant expression, we get:0.0156 = x²/(0.660 - 2x)²Solving for x, we get:x = 0.0244 MTherefore, the equilibrium concentration of HI is:[HI] = 0.660 - 2x = 0.660 - 2(0.0244) = 0.244 M

Hydrogen Iodide Decomposes According to the Equation: 2HI(g) ⇌ H2(g)

Hydrogen Iodide Decomposes According to the Equation: 2HI(g) ⇌ H₂(g)