In this problem we are going to be reacting C2H4 and C2H4 is a gas. It doesn't say up there but I'm letting you know it's a gas. It's reacting with hydrogen. Now we know that that's a gas. To produce C2H6. This too is a gas. Now in your past, you may have learned about Avogadro's law. You should have learned about Avogadro's law. And a constant temperature and pressure, we know that we can think about the balanced equation, not in terms of just moles, but also in terms of volume units. So, if you have 100 millilitres of this substance reacting with 100 millilitres of this substance, it's equal amounts, and you're going to, according to the 1 to 1 to 1 ratio, you would produce 100 millilitres of this substance. And that's because it's at constant temperature and pressure. Now we're going to calculate the amount of work done in this reaction. When a gas is involved, work can be calculated by this equation. Work is equal minus p times delta v, where p is the external pressure of 1.5 atmostpheres. Delta V is a change of volume of final minus initial. Now, I have got a conversion factor between the work unit or the energy unit of liter atmospheres to joules. I know how to go between those two. So I'm not going to put millilitres into this equation. I'm going to put it in terms of liters. So final that's product. I have 100 millimeters of gas. That would be 0.1 liters. And I'm, initially started with 200 millilitres, or 0.2 litres. Now, when you multiply this out, you're going to get 0.15 liter atmospheres. So that is a work unit that is an energy unit. And it's going to be, since it's an a, it's 0.1 minus 0.2. So, this is going to be negative and I changed the signs. So, we have positive work. Positive work is always what happens when a gas compresses in, gets smaller. So the work is being done by the surroundings on the system, and that is a positive work. And it works out that way with our equation as well. But we don't want to leave it as liter atmospheres because it asks for it in joules. So I've got a relationship between liter atmospheres and joules. There are 101.3 joules in a liter atmosphere. Now the two significant figures, that is 15 joules. So that's how much work is done as this reaction takes place.