Use standard enthalpies of formation to calculate ΔH °_rxn for each reaction. a. C₂H₄( g) + H₂( g) → C₂H₆( g)

Video transcript

hey everyone today, we're being asked to find the standard entropy of the reaction for the reaction of phosphorus becoming di phosphorus and tetra phosphorous at equilibrium. So to do this, we need to utilize the standard entropy of formation in order to find the standard entropy of the reaction as a whole. And we can use a general formula that goes as such. The standard entropy of the reaction reaction is equal to the difference between the standard entropy of the formation of the products and the standard entropy of formation of the reactant reactant. So with that in mind we actually need to find the specific values for the standard entropy of formation for each of the compounds present in the reaction. So let's go ahead and take a look at phosphorus first. So phosphorus, the standard entropy of formation, which is a experimentally calculated value and can be found in tables either online or in a textbook, but for phosphorus specifically phosphorus gas, The entropy of formation is 316. Kill it jules per mole for di phosphorus down to h the standard entropy of formation is 144 killer jewels per mole. And for tetra phosphorous, a standard entropy of formation is 58 0.9 kg joules per mole. So putting this into our products products minus reaction formula and let's keep our products in blue. We also need to consider how many moles of everything we have. So for the products we only have one mole produced of each. So we have one mole times 144 Kill it, jules per mole plus uh, one more Times 58.9 kg per mole. So this will be the standard entropy of formation for the products. And from here we will subtract the standard entropy of formation of the reactant. And we have six moles of loan phosphorus, that's what we have right here. So we actually factor that in as well. So we have six moles of p Times the 316.5 kila jules per mole and adding it all up. We get a final value. The final standard entropy of the reaction of negative 1696.1 kg jewels. I hope this helps. And I look forward to seeing you all in the next one.