Be sure to answer all parts The dissociation of molecular io

Be sure to answer all parts. The dissociation of molecular iodine into iodine atoms is represented as I2(g) 2I(g) At 1000 K, the equilibrium constant Kc for the reaction is 3.80 × 105. Suppose you start with 0.0453 mol of I2 in a 2.28L flask at 1000 K. What are the concentrations of the gases at equilibrium? What is the equilibrium concentration of I2? M What is the equilibrium concentration of I? M

Solution

Equilibrium constant (KC) for the reaction I2<----> 2I

is

Kc= [I]²/ [I2]

Given initial concentration of I2= moles/liter = 0.0453/2.28 M=0.0198M

Let x= drop in concentration of I2 to reach equilibrium

At equilibrium, [I2]=0.0198-x and [I]= 2x

So Kc= 3.80*10^-5 = (2x)²/ (0.0198-x)

Hence x²/(0.0198-x)= 9.5*10^-6, when solved using excel , x= 4.3*10^-4M

At Equilibrium [I2] =0.0198-4.3*10^-4 =0.01937M and

[I]= 2*4.3*10^-4 = 8.6*10^-4M