At 340. K, K_P = 69 for the reaction H₂(g) + I₂(g) 2HI(g). Suppose 50.0 g of HI is injected into an evacuated 5.00-L rigid cylinder at 340. K. What is the total pressure inside the cylinder when the system comes to equilibrium?

Question

Quizplus · Accepted Answer

The balanced chemical equation is: H₂(g) + I₂(g) ↔ 2HI(g) The reaction quotient (Qp) can be expressed as: Qp = P_P / (PHI)2 At equilibrium, Qp = K_P Therefore, K_P = PHI2 / (P_P) We can assume that the initial pressure of HI is so low that it can be neglected in the equilibrium expression. Therefore, K_P = (Ptotal - PHI)2 / (Ptotal) At equilibrium, the total pressure (Ptotal) inside the cylinder is equal to the sum of the partial pressures of all the gases present. Let x be the equilibrium partial pressure of HI. Then, the equilibrium partial pressures of H₂ and I₂ are both equal to (0.050 - x) g/L. Substituting the values into the equilibrium expression and solving for x, we get: 69 = (5.00 - x)2 / [(0.050 - x)2] Solving for x using quadratic formula, we get: x = 0.474 g/L The total pressure at equilibrium is: Ptotal = x + 2(0.050 - x) = 0.15 atm Converting to atm, we get: Ptotal = 2.18 atm Therefore, the best choice is D.

At 340 K, KP = 69 for the Reaction H2(g)

At 340 K, K_P = 69 for the Reaction H₂(g)