It is interesting to compare the 'fuel value' of a hydrocarbon in a hypothetical world where oxygen is not the combustion agent. The enthalpy of formation of CF₄(g) is -679.9 kJ/mol. Which of the following two reactions is the more exothermic?

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)

CH₄(g) + 4 F₂(g) → CF₄(g) + 4 HF(g)

Video transcript

Hello. Everyone in this video. We're given these two actions and is saying that the hydrocarbons can be burned using cl two gas as well as 02 gas. So using the 02 gas, would this be less or more of an exit thermic reaction? And we're using these two equations to go ahead and do a comparison. So first things first is I'm going to go ahead and calculate the delta H of the reaction for both equations. And why is that is because we can see by the sign or the just the values of the delta H of the reaction. If it's going to be more exact thermic or less exotic thermic and that's going to go ahead and answer this question right here. And so there's going to be a formula that we can use for the delta H over reaction. I'll go and write this in black. So the delta H of this reaction is going to equal to the delta H. Of all the products. So the sum of all adult age of the products minus the delta age of our products. So reframing the some of this, subtracting it by the some of this for delta H. S. All right, so label this as A and the second as B. So or eight also for the delta H values of the products and the um oh I realized I just wrote products twice. So this should be reactant. So delta age of reaction. Is that to age of the products minus delta age of our reactant. So these values can be found either in your textbook are given to you by a professor, I'm going to be using the values that so continuing on for the first one, we have that detail to H of the reaction is going to equal to -393.5. Kill jules per. And we're going to go ahead and multiply this by three moles because we have three moles of a C. 02. Okay then we're gonna go ahead and add The Delta H for the H. two. And that's going to be negative. 241.82 kilo jewels. Her more are going to multiply this by formals. Alright, and this is going to be subtracted by the delta ages of our reactant In our case. First we have our c. H8 On the Delta H. For that is going to be negative 103 0. killer jewels. Permal times. We have just one more of this. And then any delta age regarding just an element in its natural state of course are owed two cl two. They're both di atomic molecules or Di Atomic atoms and that's why in its natural state which is in this gaseous phase, that's going to be 02 and cl two. And that delta H is going to be all zeros for all natural elements. All right, so we'll just have adding zero. Now putting all of this into my calculator. The Delta Age of the reaction for eight, It's going to be - 93 Killah jewels per Now doing this again. This whole process for girlhood and scroll down. So see here we have again the delta H. Of the reaction. You going to first we'll be doing the delta H. S. For our CCL four. That's going to have a delta age of negative one or 6.7 killer jules Permal. And we're going to go ahead and multiply this by three malls. And then we're gonna go ahead and add delta H. For the H. C. L. And that's going to have adults age of negative 92. kila jules Permal times eight moles. And that's going to be subtracted by they delta age of our starting materials. And of course we have only delta age of our C. Three H eight because cl two is a di atomic molecules. And so We have a delta h of that is going to be negative 103.85 kila jules. Permal times one more plus zero. Now putting all that into my calculator, I'll get a delta H reaction equaling two -954 0. 65 killer jewels, her mom. Now if you compare this value and this value. This delta age reaction for this second equation here, it's going to be a lot more positive and less negative. And because of that, we know that using the cl two to go ahead and burn is going to be less eggs are thermic process. So the answer to this question, it's going to be burning C L two gas is going to be less exhaust thermic, and that is going to be my final answer for this question.