Using data from Appendix C, calculate ΔG° for the following reactions. Indicate whether each reaction is spontaneous at 298 K under standard conditions. (a) 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Video transcript

Welcome back everyone in this example, we need to determine the standard change in gibbs Free energy and spontaneity for the given reaction at 298 kelvin. Under standard conditions were given the standard gibbs free energy of formation of nitrogen tetroxide, nitrogen monoxide and of oxygen gas. And were given our bouncy equation below here. So recall that to calculate our standard gibbs free energy change, we take our standard gibbs free energy of formation of our products subtracted from the standard gibbs. Free energy of formation of our reactant. And because we're given all of this info above, we can just plug in what we know, we would say that and I'll just scoot this over for more room. Actually that our standard gift free energy of our reaction is equal to beginning with our only product, we have di nitrogen tetroxide. So we have a or we have one mole of that, Which is multiplied by its standard Gibbs free energy of formation of a value being 99.8 kg joules per mole as given in the prompt. So this completes our some of our standard gibbs. Free energy of formation of our products. And now we're going to subtract from the standard gibbs. Free energy of formation some of our reactant where we begin with our first reactant which is our two moles of nitrogen monoxide, multiplied by its standard energy. A formation given in the prompt as 87.6 kg per mole. We continue our brackets below in the next line where we have this added onto our two or sorry, one mole of oxygen gas multiplied by oxygen gasses. Standard gibbs free energy of formation given in the prompt as zero killer jewels per mole, completing our some of our gibbs. Free energy of formation of our reactant. Notice we can cancel our units of moles because it's aligned in the numerator and denominator leaving us with units of killing jewels as our standard gibbs. Free energy um Change for our reaction. So simplifying this for the multiplication of our products. In our brackets here we have a value of 99.8 kg joules, Which is being subtracted from our brackets here, which we can simplify to 175. kg jewels per mole. And so this would give us our standard change of Gibbs. Free energy for reaction equal to the difference, which gives us a value of negative 75.4 kg jewels. And sorry, this should not have the units of kilograms because as we did above, we canceled the units of moles out recall that when we have a negative value for gibbs, free energy change that corresponds to a spontaneous reaction. But when we have a positive value for gibbs, free energy change that will correspond to a non spontaneous reaction. Now, in this case we have a negative value, meaning that we can say therefore our reaction is spontaneous. And so for our two final answers, we've calculated the standard free energy change of our reaction as negative 75.4 kg jewels, and we determined that based on this negative value of our gibbs, free energy change, the reaction is spontaneous. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.