Answer to Question #100387 in Physical Chemistry for Gaby

Question

For the reaction: H2 (g) + I2 (g) <--> 2HI (g)
Keq= 50.0 at 448°C

How many moles of HI are present at equilibrium when 2.0 mol of H2 react with 1.0 mol of I2 in a 0.50L flask at 448°C?

Answer 1

In this problem we are given moles of H₂ and I₂ , so we will find out their their concentration as in equilibrium we deal with concentration

as, concentration = moles /volume

concentration of H₂ = 2/0.5 = 4 mol/L

concentration of I₂ = 1/0.5 = 2 mol/L

the given reaction is :

 H_{2 (}g) + I₂ (g) <--> 2HI (g) 

initial 4 2 0

final 4-x 2-x 2x .......(1)

equilibrium constant(K) = [HI]² /[H_2]][I₂]

50 = (2x)²/(4-x)(2-x)

50(x² -6x+8)=4x²

46x²-300x+400=0

on solving the quadratic we get x=4.7 and 1.9

4.7 is large as we have started with 4 mol/L of H₂

hence x= 1.9

substituting the value of x in the equation (1)

we get [H₂]=4-1.9=2.1 mol/L

[I₂]=2-1.9=0.1 mol/L

[HI]=2*1.9=3.8 mol/L

concentration of HI is 3.8

hence number of moles will be = concentration * volume = 3.8*0.5= 1.9 moles