# Question #00c38

##### 2 Answers

Lets assume that we start with 2 moles of steam:

We are told that 20% of the steam dissociates. This means that 20% of 2 moles dissociate which = 2 x 0.2 = 0.4 moles.

This means that the number of moles of steam left @ equilibrium = (2 - 0.4) = 1.6 mol

From the equation we can see that if 0.4 mol steam dissociate then 0.4 mol of hydrogen must form as well as 0.2 mol oxygen.

So equilibrium moles

Which is:

So the total moles @equilibrium

We can now work out the mole fraction of each gas since

We can then find the partial pressure of each gas since for a gas

Lets do a check here and see if they add up to the total pressure of

Now for

**!! LONG ANSWER !!**

Here's how you should approach this problem.

You know that at a *total pressure* of **1 atm** and at very high temperatures, water dissociates to form hydrogen gas,

The equilibrium reaction that describes that dissociation looks like this

Notice that **2 moles** of water produce **2 moles** of hydrogen gas and **1 mole** of oxygen.

You also know that only **20%** of the molecules dissociate. If you have *total number of molecules* present, then you have

*the number of water molecules that dissociate*;

*the number of water molecules that don't dissociate*.

Out of the molecules that do dissociate, you know that each pair produces a pair of hydrogen molecules and 1 oxygen molecule. This means that you have

*the number of molecules of hydrogen produced*;

*the number of molecules of oxygen produced*.

So, at equilibrium, your vessel will contain

The total number of molecules will be

You can write the partial pressure of each gas by using the **mole fraction** they hold in the mixture and the total pressure of the mixture by using **Dalton's Law of Partial Pressures**.

The mole fraction is defined as the number of moles of a gas divided by the total number of moles of gas present in a mixture. In your case, you have

The partial pressures of the three gases will be

The equilibrium constant,