Consider the following equilibrium: N₂O₄(g) ⇌ 2 NO₂(g) Thermodynamic data on these gases are given in Appendix C. You may assume that ΔH° and ΔS° do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases?

Video transcript

Hello everyone. So in this video we're calculating for the temperature of this equilibrium reaction right over here assuming that the mixture contains equal amounts of the two gasses. So first things first is to go ahead and realize that K value is equal to the pressure from our S two ratio fourth power Over our pressure of r. s. eight. Okay, so we're thinking about the gas law equation of So let's write that in different color. Maybe. So the gas law is equal to our pressure equaling two N. R. T over V. So we're being told that the R. T over V value is constant because the reaction is at equilibrium and at equilibrium there's going to be equal partial pressure. I'll just put the Spp which means that we have equal moles of gas. Okay, so the value of K will then depend on the total pressure of the mixture. And we here assuming that. Well, because it is said in the statement that our pressure of S two equals to the pressure of R. S eight and that is a value 1.0 A. T. M. Therefore RP total then would be these two added up. So these two values hold each a value of 1.0. The total will be 2.0 A. T. M. So go ahead and plug this into our K. Constant equation to give me that K is equal to. Well, let's see here we have 1.0 raised to the fourth power over 1.0. So point that into my calculator. I will get that mic a constant value is equal to 1.0. And we have when we have that the K. Is equal to one. Our delta G. Will be equal to zero. Let's go ahead and scroll down. So we're using our gifts free energy equation Now to solve for the temperature, that's going to be that the delta G. Is equal to delta H minus temperature, multiplied by delta S. Where rearranging this equation to software T. So we get that T. Is equal to our delta H. Over our delta S. So let's first solve for each component of this thi equation. So first let's do delta H. So that is of delta H. Of our products. So just put that is p minus delta H. Of our reactant which I just put as art. So that's equal to four times 28.6 minus one oh 2.3. So once I put that into my calculator I will get the value of 412.1 kg joules. Now for our delta us same exact thing. So that value of our products minus of the reactant. So that being four times 2 to 8.2 minus 430.9. Again putting this into my calculator, I'll get the value of 481.9 joules per kelvin. So now that we have solved for each component, we can now then plug everything into our T. Equation. So T. Eagles while we have the value 412.1 killed jewels for our delta H. We want that in terms of jewels. So for every 1000 jewels we have one kill a jewel. So that's our numerator are delta s. That's going to be 481 0. jewels per kelvin. So we can see here that these values council and the jewels will cancel leaving us with the value that T. Is equal to 855. kelvin's. And this right here is going to be my final answer for this problem.