Hydrogen iodide decomposes according to the equation 2HI(g) H₂(g)+ I₂(g), for which K_c = 0.0156 at 400ºC. 0.550 mol HI was injected into a 2.00 L reaction vessel at 400ºC.Calculate the concentration of HI at equilibrium.

Question

Quizplus · Accepted Answer

First, we need to use an ICE table to determine the equilibrium concentrations of all species: | | 2HI | H2 | I2 | |--------|--------|--------|--------| | Initial| 0.550 M| 0 | 0 | |Change | -2x | x | x | | Equil. | 0.550 - 2x| x | x | Using the equilibrium constant expression and plugging in the equilibrium concentrations, we get: K_c = [H2][I2]/[HI]^2 0.0156 = x^2/(0.550 - 2x)^2 Solving for x (the change in concentration of HI), we get x = 0.165 M. Therefore, the equilibrium concentration of HI is: [HI] = 0.550 - 2x = 0.220 M. So the answer is choice B.

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