Answer to Question #107097 in General Chemistry for mya 8

Question

At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3.

H2(g)+I2(g)−⇀2HI(g)Kc=53.3

At this temperature, 0.500mol H2 and 0.500 mol I2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?

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

$HI=2\times0.392=0.785$