Question

A certain quantity of HI was placed in a 1.00 L container, when equilibrium is reached [I2]=1.8 and kc= 6., what was the original number of moles of Hi placed in the reaction flask?

***please show all work... i want to understand how to do it

Answer 1

The equilibrium can be represented as

2 HI <=====> H₂ + I₂

The equilibrium constant for the reaction is given as

K_c = [H₂][I₂]/[HI]² = 6 where the square braces denoted molar concentrations.

Set up the ICE chart. Let us start with a moles of HI and let x moles of HI be converted to H₂ and I₂ at equilibrium. Since the volume of the reaction vessel is 1.00 L, we can replace the molar concentrations by the number of moles (since molar concentration = number of moles/volume of container).

2 HI <=======> H₂ + I₂

initial                                                    a                         0      0

change                                               -2x                     +x    +x

equilibrium                                      a – 2x                      x      x

Why we have the change in number of moles of HI as -2x? Since we have x moles each of H₂ and I₂ forming, we employ the stoichiometric equation to find out that the change in number of moles of HI is -2x.

As per the problem, [H₂] = [I₂] = (x mole)/(1.00 L)

Therefore,

(x mole/1.00 L) = 1.8 M = 1.8 mole/L

===> x = 1.8

Therefore, [H₂] = [I₂] = (1.8 mole)/(1.00 L) = 1.8 M.

Plug in the expression for K_c and obtain

6 = (1.8)*(1.8)/(a – 2*1.8)²

===> 6 = 3.24/(a – 3.6)²

===> (a – 3.6)² = 3.24/6.00 = 0.54

Take positive square root on both sides,

a – 3.6 = 0.7348

===> a = 3.6 + 0.7348 = 4.3348 ≈ 4.3

The original concentration of HI is 4.3 mol/L; therefore, the original number of moles of HI is (4.3 mol/L)*(1.00 L) = 4.3 moles (ans).