Problem

Using the standard enthalpy of formation data in …

05:42

Problem 70 Medium Difficulty

Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbon-sulfur double bond in CS2

Answer

578.7 $\mathrm{kJ} \cdot \mathrm{mol}^{-1}$

Allea C.

University of Maryland - University College

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Video Transcript

although some of the concepts associated with solving this problem we're not introduced in the chapter. They were introduced in problems 69 so you can review problems 69. And it's answer in the back of the book in order to reiterate these concepts that I will be introducing. First, we need to recognize that the bond energy for the double bond and C s to is the energy that is required to break apart, um, or of the CS two molecule into one mole of gaseous carbon and two moles of gaseous sulfur atoms. Just Adams. So when I say gaseous carbon, I'm talking about gaseous carbon atoms. We also need to know that the energy associated with the formation of CS two, which is called the Standard and Thall P of formation, is carbon in its standard state, which is graphite as a solid plus sulfur in its standard state with which is s eight solid in a ram. Bic geometry is defined by the reverse of this reaction. So if we start with CS two and break it apart into these standard forms, then this would be the reverse of the Delta H of formation of CS to gas. So the Delta H for this reaction as written is going to be the negative Delta H of formation of CS two. This will be the first chemical reaction that we're considering if we then define the energy for carbon solid as a graphite being broken apart into gaseous carbon. The Delta H for this process is going to simply be the Delta H of formation of gaseous carbon atoms. Again, Delta H of formation is taking the element in its standard state and making what we're interested in. One mole of what we're interested in so carbon in its standard state is graphite. We will then be making one mole of gaseous carbon atoms. We could do the same thing with sulfur sulfur. In its standard state is s eight solid Rumbek. This can then be broken apart into sulfur atoms. And, um, normally, Delta H of formation is simply one mole of the product we're interested in. But it's going to be helpful for what I'm going to do next to form two moles of sulfur gas because there's two moles of sulfur for every one mole of carbon. So this chemical reaction then would have a Delta eight that is equal to two times the delta H of formation of of a mole of sulfur atoms in the gas phase. Now, the reason I chose thes three chemical reactions with their associative Delta H value is because when I sum them up, they're going to give me the chemical reaction associated with the Bond Energy of CS in CS two, you will notice that the 1/4 s a s rahm Bix will cancel and the carbon graphite will cancel. And I will get a chemical reaction showing gaseous CS two being broken apart into its atoms, carbon atoms and to sulfur atoms. This is the chemical reaction that is defining the bond energy of CS in CS two More specifically, the Delta H for this process is going to be two times the bond energy because when this process occurs to C. S, double bonds are broken and the Delta H for this process is simply going to be the some of the Delta. H is for the processes above, because this reaction was the some of these chemical reactions. So if we look up Delta H of formation of CS two, we get positive. 1 16.9. The negative of that would be negative. 1 16.9 Delta H of formation of carbon is 7 16.6 and then the Delta H of formation of sulfur Adams's to 78 killer jewels. But we want to times that to be equal to Delta H three. So summing up Delta H one, Delta H two and Delta H three, we get two times the bond energy of C Double Bond s, which is one 1001 157 0.401 killer jewels Permal. But that is two times the bond energy. So we're going to have to divide that by two in order to get the energy associated with the breaking of just one mall of C double Bond s bonds, which would be 5 78.7 killer jewels

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