Question

Consider the reaction: H2(g)+I2(g)⇌2HI(g)

A reaction mixture in a 3.64 −L flask at 500 K initially contains 0.376 g H2 and 17.97 g I2. At equilibrium, the flask contains 17.76 g HI.

Part A Calculate the equilibrium constant at this temperature.                I keep getting 13413.06 and its not right I'm running out of tries, please help me.

Answer 1

M = mol per liter

claculate all to mol, then divide by L to get concnetrations

K = [HI]^2 / [H2][I2]

mol H2 = 0.376/2 = 0.188

mol I2 = 17.97/253.809 = 0.0708012

mol HI = 0

thne

[I2] = mol/V = 0.188/3.64 = 0.051648

[H2] = mol/v = 0.0708012/3.64 = 0.0194508

[HI] = 0

in equilibrium

[I2] =0.051648 - x

[H2] = 0.0194508 - x

[HI] = 0 + 2x

since

[HI] = mol/V = mass/MWL = 17.76/(3.64*127.911 ) = 0.038144

[HI] = 0 + 2x =0.038144

x = 0.038144/2 = 0.019072

then

[I2] =0.051648 - x = 0.051648 -0.019072 = 0.032576

[H2] = 0.0194508 - x = 0.0194508  -0.019072 = 0.0003788

[HI] = 0 + 2x = 0.038144

K = (0.038144)^2 /(0.0003788*0.032576) = 117.908