Calculate ΔHrxn for the reaction: CH4( g) + 4 Cl2( g)¡CCl4( g) + 4 HCl( g) Use the following reactions and given ΔH's: C(s) + 2 H2( g)¡CH4( g) ΔH = -74.6 kJ C(s) + 2 Cl2( g)¡CCl4( g) ΔH = -95.7 kJ H2( g) + Cl2( g)¡2 HCl( g) ΔH = -92.3 kJ

Video transcript

Hello, everyone. Today we have the following problem. Calculate the ed for the reaction, use the following reactions and given changes in entropy. So we can use what is known as Hess's law that states some of the entropies of individual steps is equal to the entropy of the overall reaction. So let's look at reaction one, we see here that when compared to the overall reaction, we have our methane on the product side and our first reaction. But in our overall reaction, it is on the reactants. So we essentially have to reverse our sign of our entropy change. So that goes from an entropy of a negative 74.6 to positive 74.6 kilojoules. We look at our second reaction, we see that our fluorine gas is on the reactant side and the overall reaction and so is our carbon tetra fluoride that's on the product side. So there is no change. We then look at our third and final reaction. And we see here that while we do have our Hydrofluoric acid on the product side, we see that is or it has been halved. So what we need to do is we need to multiply our entropy change by that variable or two such that our new entropy will be negative 1084 kilojoules. And then to solve for hour entropy of the reaction, the total reaction, we simply add up the individual entropies. So we have our positive 74.6 kilojoules that will get added to our negative 933 kilojoules. And then lastly, we add by our negative 1084 kilojoules for a final value of negative 1942.4 kilojoules or WA C. And with that, we have solved the problem overall, I hope is helped. And until next time.