At 1400 K, the value of K c for 2 HBr (g) H? (g) + Br?(g) is 1.5*10 - 5 . What is the equilibrium concentration of H 2 in a 1.00 liter container that initially held 0.10 mol HBr and 0.15 mol of Br?

Question

At 1400 K, the value of K c for 2 HBr (g)  H? (g) + Br?(g) is 1.5*10 - 5 . What is the equilibrium concentration of H 2 in a 1.00 liter container that initially held 0.10 mol HBr and 0.15 mol of Br?

Quizplus · Accepted Answer

The equilibrium concentration of H₂ can be found by using the equilibrium expression:
Kc = [H₂]¹¹[Br₂]/[HBr]²
Let x be the equilibrium concentration of H₂. Then, the equilibrium concentrations of HBr and Br₂ will be (0.1 - 11x) and (0.15 - x), respectively.
Substituting these values into the equilibrium expression and solving for x gives:
1.5×10⁻⁵ = x¹¹(0.15 - x)/(0.1 - 11x)²
Solving this equation using a numerical method (such as a graphing calculator or computer software) gives x = 1.0×10⁻⁶ M.
Therefore, the equilibrium concentration of H₂ is 1.0×10⁻⁶ M.

At 1400 K, the Value of K C for 2