# In a 10 litre evacuated chamber, 0.5 mole of H2 and 0.5 mole I2 gases are reacted to produce HI gas at 445°.the equilibrium constant at this temperature is 50. How many moles of iodine remain unreacted at equilibrium?

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

Before doing any calculations, try to predict what you expect to see once the reaction reaches equilibrium.

Take a look at the value of the *equilibrium constant*,

The fact that you have **forward reaction**, i.e. the *consumption* of the reactants and the *formation* of the product.

The *bigger* the value of *very small* in comparison with that of the product.

Now, use the volume of the chamber and the number of moles of each reactant to calculate their initial concentrations

#color(blue)(c = n/V)#

#["H"_2] = "0.5 moles"/"10 L" = "0.05 M"#

#["I"_2] = "0.5 moles"/"10 L" = "0.05 M"#

You can now use an **ICE table** to help you find the equilibrium concentrations of the three species

#" " "H"_text(2(g]) " "+" " " " "I"_text(2(g]) " "rightleftharpoons" " color(red)(2)"HI"_text((g])#

By definition, the equilibrium constant for this reaction will be equal to

#K_c = ["HI"]^color(red)(2)/( ["H"_2] * ["I"_2]) = (color(red)(2)x)^color(red)(2)/( (0.050 -x )(0.050 -x))#

This is equivalent to

#K_c = (4x^2)/(0.050 - x)^2 = 50#

Take the square root of both sides to get

#sqrt((4x^2)/(0.050 - x)^2) = sqrt(50)#

#(2x)/(0.050 -x ) = sqrt(50)#

Solve this for

#2x = 0.050 * sqrt(50) - x * sqrt(50)#

#x * (2 + sqrt(50)) = 0.050 * sqrt(50)#

#x = (0.050 * sqrt(50))/(2 + sqrt(50)) = 0.0390#

This means that the equilibrium concentrations for the species involved in this reaction will be

#["H"_2] = "0.050 M" - "0.0390 M" = "0.011 M"#

#["I"_2] = "0.050 M" - "0.0390 M" = "0.011 M"#

#["HI"] = 2 * "0.0390 M" = "0.078 M"#

As predicted, the equilibrium concentration of *hydrogen iodide*, **bigger** than the equilibrium concentrations of the two reactants.

Finally, to get the number of moles of iodine,

#color(blue)(c = n/V implies n = c * V)#

This will get you

#n = 0.011 "moles"/color(red)(cancel(color(black)("L"))) * 10 color(red)(cancel(color(black)("L"))) = color(green)("0.11 moles I"_2)#

I'll leave the answer rounded to two sig figs.