An equilibrium mixture of H2, I2, and HI at 458 degrees Celsius contains 0.0224

Question

An equilibrium mixture of H2, I2, and HI at 458 degrees Celsius contains 0.0224 M H2, 0.0224 M I2, and 0.155 M HI in a 5.00 L vessel. What are the equilibrium concentrations when equilibrium is reestablished following the addition of 0.100 mol HI?

Explanation / Answer

H2 + I2 -------> 2HI Kc = 0.155^2 / 0.0224 * 0.0224 = 47.88 After 0.1 moles HI is added [HI] = 0.155 + 0.1 / 5 = 0.175 M H2 + I2 ------> 2HI 0.0224+X;0.0224+X; 0.175-2X Kc = 47.88 = (0.175-2X)^2 / (0.0224+X)(0.0224+X) => 0.175-2X / 0.0224+X = 6.92 => X = 0.00224 Therefore Reestablished equilibrium conditions are [HI] = 0.17052 M [H2] = [I2] = 0.02464 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---