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 cC + dD the K_c 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 since their composition does not change throughout the reaction. The standard state of a pure substance is the pure substance itself, and although the quantity may change the sample remain pure. The concentration is effectively equal to 1, and will not impact the magnitude of K Deriving concentrations from data 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 723 degree C: H_2(g) + I_2(g) 2HI(g) Initially, only H_2 and I_2 were present at concentrations of [H_2] = 3.90M and [I_2] = 2.50M The equilibrium concentration of I_2 is 0.0700 M. What is the equilibrium constant. K_c, for the reaction at this temperature? Express your answer numerically.

Explanation / Answer

H2 (g) + I2 (g)   ßà 2HI (g)

3.90        2.50         0     (initial)

3.90-x      2.50-x        2x    (at equilbrium)

at equilibrium,

[I2] = 2.50-x = 0.0700 M

x = 2.43 M

Kc= [HI]^2 / {[H2][I2]}

   = (2x)^2 / {(3.90-x)(2.50-x)}

   = (2*2.43)^2 / {(3.90-2.43)(2.50-2.43)}

   = 229.5

Answer: 229.5

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