Phosphorus pentachloride decomposes according to the chemical equation pcl 5 ( g ) − ⇀ ↽ − pcl 3 ( g ) + cl 2 ( g ) k c = 1.80 at 250 ∘ c pcl 5 ( g ) ↽ − − ⇀ pcl 3 ( g ) + cl 2 ( g ) k c = 1.80 at 250 ∘ c a 0.342 mol 0.342 mol sample of pcl 5 ( g ) pcl 5 ( g ) is injected into an empty 3.65 l 3.65 l reaction vessel held at 250 ∘ c. 250 ∘ c. calculate the concentrations of pcl 5 ( g ) pcl 5 ( g ) and pcl 3 ( g ) pcl 3 ( g ) at equilibrium.

See explanation for answer

Explanation:

First, let's write the equation again:

PCl5 <> PCl3 + Cl2      Kc = 1.8

The expression for equilibrium is the following:

Kc = [PCl3] [Cl2] / [PCl5]

We only know the data for PCl5 at the beggining, and to solve this, we need to write what happens before and after the reaction.

At first, we only have the 0.342 moles of PCl5 and nothing of the products. This is logic, because there is no reaction yet.

Concentration of PCl5 at this point (With a volume of 3.65 L) is:

[PCl5] = 0.342 / 3.65 = 0.094 M

Keep in mind, that this is not the concentration in equlibrium. This is only the concentration at the beggining. To know it's concentration in equilibrium, we'll do the following:

       PCl5 <> PCl3 + Cl2      Kc = 1.8

i)      0.094                  0        0

eq)   0.094 - x             x        x

Now, we will replace this values in the equilibrium equation:

1.8 = x * x / 0.094 - x

Solving for x we have:

1.8(0.094 - x) = x²

0.1692 - 1.8x = x²

x² + 1.8x - 0.1692 = 0

Using the general equation for x we have:

x = -1.8 ± √(1.8)² - 4 * 1 * (-0.1692) / 2

x = -1.8 ± √3.9168 / 2

x = -1.8 ± 1.98 / 2

x1 = -1.8 + 1.98 / 2 = 0.09

x2 = -1.8 - 1.98 / 2 = -1.89

This means that the value of x is 0.09 therefore, the concentrations in equilibrium for each species is:

[PCl5] = 0.094 - 0.09 = 0.04 M

[PCl3] = [Cl2] = 0.09 M

- both b and d.

-since b and d are the closest coastlines then that makes it the most affected.

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