The reaction S81g2 S 4 S21g2 has ΔH° = + 237 kJ (b) The average S ¬ S bond dissociation energy is 225 kJ/mol. Using the value of ΔH° given above, what is the S ' S double bond energy in S21g2?

Video transcript

Welcome back everyone Consider the following reaction. We have touch of phosphorus gas which associates into two moles of di phosphorus gas with an entropy value of 2 29.1 kill equals if the phosphorus phosphorus pond has an average bond association energy of 1 97 kg joules per mole estimate the value of the phosphorus phosphorus triple bond energy and phosphorus in die phosphorus gas. We're going to begin by drawing out our molecules that represent each of our gasses in the reaction. So beginning with tetra phosphorous, we have four moles or sorry for atoms of phosphorus which are going to be single bonded to one another where we have diagonal bonds connecting our corner phosphorus atoms. And this dotted line represents our sixth bond here. So we can say that we have six moles of our phosphorus phosphorus single bond on the reactant side which produces on our product side according to the prompt. Two moles of our phosphorus phosphorus triple bond product. So two moles of our phosphorus phosphorus triple bond. Next we want to recognize that we can use our entropy of our reaction to be calculated from the difference between the bond association energy of our reactant bonds subtracted from the bond association energy of our product bonds. So we're going to just plug in our knowns given from the prompt to solve or rather isolate for the bond dissociation energy of our phosphorus phosphorus tripled on to answer this prompt. So plugging in our will be from the prompt were given a value of 29.1 kg jewels. And this is set equal to our right hand side. Where for our bond association energy of our reactant bonds. We're going to plug in that we have six moles of our phosphorus phosphorus single bond which according to the prompt, has a bond association energy of 197 kg joules per mole. This is then subtracted from our bond association energy of our product bonds. And we'll use the color purple for that. Where we have our two moles of our phosphorus phosphorus single bond that we need to solve the bond association energy for and sorry, triple bond of the phosphorus phosphorus triple bond. So we're going to simplify this and we're going to simplify by taking the product of six moles times 187. And in our next line we should have to 29.1 kill equals set equal to our product of kg joules because we'll be able to cancel out our units of moles leaving us with kayla jewels. And this is going to be subtracted from two times. Are sorry two moles times our bond association energy of our phosphorus phosphorus triple bond. So continuing to isolate we're going to Add to both sides or correction. We're going to multiply both sides by -1. As well as ad 1,182 kg jewels to both sides. And this is going to leave our variable where we have our moles multiplied by our bond association energy of our phosphorus phosphorus triple bond fully isolated, which we can now say is equal to minus to 29.1 kg jewels. And sorry, this should be 1182 kg joules minus to 29.1 kill jules. And from this difference we would simplify in our next line to to moles times our bond association energy of our phosphorus phosphorus triple bond set equal to our difference, which results in 952. illegals. And now we want to get rid of the two moles. So we're going to divide both sides by two. And this is going to give us our variable for the bond association energy of our phosphorus phosphorus triple bond, which is equal to r value of our quotient, which will be positive 476 killer jewels. And this would be our final answer to complete this example. As the bond association energy for our phosphorus phosphorus triple bond. I hope that 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