Answer to Question #188597 in General Chemistry for mike

Question

Alesia, decided to mix pure iodine and hydrogen gases and allowed 80 minutes an equilibrium to be reached. The equilibrium concentrations of iodine and hydrogen gases at 490oC were each found to be 3.00mol dm-3. These figures were confirmed by Tim, Alesia’s PhD research assistant, to be correct. 

(i). What was the equilibrium concentration of gaseous hydrogen iodide at this temperature if the equilibrium constant is 45?

(ii). What were the starting concentrations of reactants?

Answer 1

Equilibrium concentration of H2= 0.500 M

Equilibrium concentration of Il2= 0.500 M

The equilibrium constant for the reaction Kc=53.3

The equation relating the equilibrium constant to the equilibrium concentrations of the reactants and products is:

$Kc=\frac{2HI}{H2\times I2}$

Now, the volume of the container is equal to 1.00 L, so the number of moles of each reactant and their respective concentrations are interchangeable.If you take x M to be the concentration of hydrogen gas that reacts to produce hydrogen iodide, you can say that the reaction will also consume x M of iodine gas and produce 2 x M of hydrogen iodide.This is the case because the reaction consumes hydrogen gas and iodine gas in a 1:1 mole ratio and produces hydrogen iodide in a 1:2 mole ratio to both reactants.

​Therefore, a new amount of substances will be

HI=2x

H2=0.500-x

I2=0.500-x

 $53.3=\frac{2x^2}{(0.500-x)\times(0.500-x)}$

$53.3=(\frac{2x}{0.500-x})^2$

$7.3=\frac{2x}{0.500-x}$

$7.3\times(0.500-x)=2x$

3.65-7.3x=2x

3.65=9.3x

x=0.392

$HI=2\times0.392=0.785$