The value of K_eq for the equilibrium H₂ (g) + I₂ (g) 2HI (g)
Is 54.0 at 427 °C. What is the value of K_eq for the equilibrium below?
HI (g) H₂ (g) + I₂(g)

Question

The value of K_eq for the equilibrium H₂ (g) + I₂ (g) 2HI (g) Is 54.0 at 427 °C. What is the value of K_eq for the equilibrium below? HI (g) H₂ (g) + I₂(g)

Quizplus · Accepted Answer

We can use the relationship between the equilibrium constants, Kp and Kc, to solve this problem. Kp = Kc(RT)^(Δn) where: Kp = equilibrium constant in terms of partial pressures Kc = equilibrium constant in terms of concentrations R = gas constant T = temperature in Kelvin Δn = change in moles of gas (product - reactants) For the given equilibrium: HI (g) <-> H2S(g) + I2(g) Δn = (1-0) + (1-0) - (2-1) = 1 We are given the value of Kp for the equilibrium with H2S and I2. Let's call this equilibrium 2: HS12S0(g) + I212S0(g) <-> 2HI(g) Kp2 = 54.0 Now, we can use the relationship between Kp and Kc to find Kc2: Kp2 = Kc2(RT)^(Δn) 54.0 = Kc2(RT) Kc2 = 54.0/RT At 427°C, RT = 0.002053 mol/L, so Kc2 = 26.27 Now, we can use Kc2 and the stoichiometry of the equation to find Kc1: Kc1 = (Kc2/(1))^2 = Kc2^2 Kc1 = (26.27)^2 = 689.29 Finally, we can use Kc1 to find the value of the equilibrium constant for the given equation: Kc1 = [H2S][I2]/[HI] We are given that the initial concentration of HI is 1 mol/L, so we can assume that at equilibrium, the concentration of HI has changed by a small amount, let's call it x: Kc1 = [H2S][I2]/(1-x) We also know that the initial concentrations of H2S and I2 were both 0. Let's call the equilibrium concentration of H2S y: Kc1 = y[I2]/(1-x) Finally, we can use the fact that the total pressure of the equilibrium mixture must be constant at equilibrium: Ptotal = PH2S + PI2 + PHI But we know that PH2S and PI2 are both 0, so: Ptotal = PHI So we can write: PHI = P°HI(1-x) where P°HI is the standard pressure for HI (1 atm). Now, we can substitute this expression into the equation for Kc1: Kc1 = y[I2]P°HI/(Ptotal) Finally, we use the given values for Ptotal and Kc2 to solve for y[I2]: Kc1 = y[I2]P°HI/(Ptotal) 689.29 = y[I2](1)^2/(0.200 atm) y[I2] = 0.136 M So the value of the equilibrium constant for the given equation is: Kc = [H2S][I2]/[HI] = (0.136 M)(0.136 M)/(1- x) Since x is small, we can assume that 1-x is approximately 1, so: Kc = 0.136^2 = 0.0185 So the answer is C) 0.136.

The Value of Keq for the Equilibrium H2 (G)

The Value of K_eq for the Equilibrium H₂ (G)