The equilibrium constant, K, of a reaction at a particular temperature is determ

Question

The equilibrium constant, K, of a reaction at a particular temperature is determined by the concentrations or pressures of the reactants and products at equilibrium. For a gaseous reaction with the general form aA + bB rightwardsharpoonoverleftwardsharpoon cC + dD the K_e and K_p expressions are given by K_c = [C]^c [D]^d/[A]^a[B]^b K_p = (P_C)^c(P_D)^d/(P_A)^a (P_B)^b The subscript c or p indicates whether K is expressed in terms of concentrations or pressures. Equilibrium-constant expressions do not include a term for any pure solids or liquids that may be involved in the reaction. What is the equilibrium constant, K_p, of this reaction? Express your answer numerically. K_p = 0, 230 In Part A, you were given the equilibrium pressures, which could be plugged directly into the formula for K. In Part B however, you will be given initial concentrations and only one equilibrium concentration. You must use this data to find all three equilibrium concentrations before you can apply the formula for K. The following reaction was performed in a sealed vessel at 755 degree C H_2 (g) + I_2 (g) rightwardsharpoonoverleftwardsharpoon 2HI(g) Initially, only H_2 and I_2 were present at concentrations of [H_2] = 3.00 M and [I_2] = 2.00M. The equilibrium concentration of I_2 is 0.0400 M. What is the equilibrium constant, K_c, for the reaction at this temperature? Express your answer numerically. K_c = Your answer implies that the initial concentrations of H_2 and I_2 are identical.

Explanation / Answer

B)

Kc = [HI]^2/([H2][I2])

initially

[H2] = 3

[I2] = 2

[HI] = 0

in equilibrium

[H2] = 3 - x

[I2] = 2 - x

[HI] = 0 +2x

and we know

[I2] = 2 - x = 0.04

x = 2-0.04 = 1.96

substitute

[H2] = 3 - 1.96 = 1.04

[I2] = 2 - 1.96 = 0.04

[HI] = 0 +2*1.96 = 3.92

Kc = [HI]^2/([H2][I2])

substitute

Kc = (3.92 ^2)/((1.04)(0.04))

Kc = 369.384

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.