In photosynthesis, green plants convert carbon dioxide and water ...

Question

In photosynthesis, green plants convert carbon dioxide and water into glucose (C6H12O6) according to the following equation:6 CO2(g) + 6 H2O(l) → C6H12O6(aq) + 6 O2(g)Estimate ∆H for the reaction using bond dissociation energies from Table 7.1. Give your answer in kcal/mol and kJ/mol. (C6H12O6 has five C―C bonds, seven C―H bonds, seven C―O bonds, and five O―H bonds).

Accepted Answer

Welcome back, everyone. We're told that the Aceto bacter bacteria can convert ethanol into acetic acid via an aerobic process. And the overall reaction is one mole of liquid ethanol reacting with oxygen gas to form acetic acid and liquid form. And liquid water. Calculate the entropy change for the reaction and use bond association energy in the table below. Even if the species are not in gaseous form, report our answer in kill cows per mole and kloos per mole, we're given that ethanol has five carbon hydrogen bonds, one carbon carbon bond, one carbon oxygen bond and one oxygen hydrogen bond. Whereas acetic acid has three carbon hydrogen bonds, one carbon carbon bond, one carbon oxygen double bond and one carbon oxygen single bond as well as one oxygen hydrogen, single bond. In our given chart below, we're given each type of bond and their associated average bond entropy in units of chili cos per mole and kilojoules per mole. So writing out the types of bonds making up first, our reactants beginning with ethanol C two H 50 H we're given from the prompt that we have five carbon hydrogen single bonds, one carbon carbon, single bond making up our molecule, one carbon oxygen, single bond and one oxygen hydrogen, single bond making up our molecule of ethanol. And next, our second reaction reactant sorry is oxygen in which we have two oxygen atoms, which will be doubly bound to one another with two lone pairs on themselves each so on each oxygen. And we can count a total of just one of our oxygen, oxygen double bonds making up our molecule of oxygen. Next, focusing on the types of bonds making up our products. Beginning with acetic acid CH 3 Coo H we have according to the prompt three carbon hydrogen single bonds, one carbon carbon, single bond, one carbon double bond to oxygen bond, one carbon carbon bond or sorry. We already wrote that we have one carbon oxygen bond as well as one oxygen hydrogen bond. And next, focusing on our product water, we have within our water molecule which has oxygen as a central atom, two bonds to hydrogen, which are single bonds. And so we have two oxygen, hydrogen, single bonds. So now with the types of bonds outlined, we want to recall our formula to calculate the entropy change of our reaction expressed by delta H which is set equal to the sum of bond entropy of our reactants subtracted from the sum of bond entropy of our products. So beginning with the sum of bond entropy of our reactants, we open up our brackets and our reactant ethanol we can say we have as the prompt states, five moles multiplied by the bond entropy of a carbon hydrogen single bond. Then added to this, we have one mole multiplied by the bond entropy of a carbon carbon single bond added to this. We have one mole multiplied by the bond entropy of a carbon oxygen single bond. And added to this, we have one mole multiplied by the bond entropy of an oxygen hydrogen single bond. This will then be added to our second reactant where we have one mole multiplied by the bond entropy of an oxygen oxygen double bond, which will then close off our brackets and complete the sum of bond entropy of our reactants. Now, beginning with the sum of bond entropy of our products, we would subtract and open up our brackets so that we have three moles multiplied by the bond entropy of a carbon hydrogen single bond. Then added to this, we have one mole multiplied by the bond entropy of a carbon oxygen double bond within ethanol added to this. We have one mole multiplied by the bond entropy of a carbon carbon single bond moving forward. We have added to this um one mole multiplied by our bond entropy of a carbon oxygen single bond. And then added to this, we have one mole multiplied by the bond entropy of an oxygen hydrogen, single bond in our structure of acetic acid. Next, we have our second product where we would add two moles multiplied by the bond entropy of an oxygen hydrogen, single bond, which would close off our brackets and complete the sum of bond entropy of our products. So now with this all outlined, let's plug in our values given in our prompt beginning with the sum of bond entropy of our reactants. We have five moles multiplied by the bond entropy of a carbon hydrogen single bond given in our prompt As kill a cows Permal. So we'll start off with our kill cow Permal and change reaction. So then this is added to our one mole multiplied by the bond entropy of a carbon carbon single bond given as 83 kg cows per mole, which is then added to one mole multiplied by the bond entropy of a carbon oxygen single bond given as 86 kg per mole, which is then added to one mole multiplied by the bond entropy of an oxygen, hydrogen single bonds given us 112 kg cows per mole. And then we have just one more type of bond for our second reactant oxygen being our one mole multiplied by the bond entropy of an oxygen oxygen double bond given us 119 kg cos per mole. So closing off our brackets and completing the sum of bond entropy of our reactants, we will now subtract from the sum of bond entropy of our products in which we have three moles multiplied by the bon entropy of a carbon hydrogen single bond given in our pump as 99 kg cows per mole added to this, we have one mole multiplied by the bond entropy of a carbon oxygen double bond given to us as 178 kg per mole added to this. We have one more which is multiplied by the bond entropy of a carbon carbon single bond given as 83 kg per mole. And added to this, we have one mole multiplied by the bond entropy of a carbon oxygen single bond given as 86 kg per mole. And added to this, we have one mole multiplied by the bond entropy of an oxygen hydrogen single bond given as 112 kg per mole, which is then finally added to our two moles multiplied by the bond entropy of an oxygen hydrogen single bond given as again, 112 kg cows per mole for our product water. Now, we're going to complete the sum of bond entropy of our products and close off our brackets. And in our next line, we would have the result of our sum of bond entropy of our reactants in our brackets being 895 kg after we cancel our unit of moles. So that will then be subtracted from the result of our sum of fun entropy of our product in brackets which is 980 kg after we cancel out our unit of moles as well. So from this difference in our next line, we would find our entropy change of our reaction in kill cows per mole as negative 85 kills. And this is going to be interpreted as per mole of reaction. So we have introduced our units of moles. Again, this would be our first answer so far as the entropy change of our reaction in Killick house per mole. And next, we want to find the entropy change of our reaction and kloos Permal. And so this is set equal to, again, the sum of bond energies of our reactants in kilojoules per mole minus are sum of bond energies of our products in kilojoules per mole. So beginning with our sum of bond entropy of our reactants, we would open up our brackets and we have our five moles multiplied by the bond entropy of a carbon hydrogen single bond then added to our one mole multiplied by the bond entropy of a carbon carbon single bond. Then added to this our one mole multiplied by the bond entropy of a carbon oxygen single bond added to this our one mole multiplied by the bond entropy of a oxygen hydrogen single bond. And then added to this. For our oxygen molecule, we have one mole multiplied by the bond entropy of an oxygen oxygen double bond, completing our brackets and completing the sum of bond entropy of our reactants. We will then subtract from the sum of bond entropy of our products in kilojoules per mole. So that we have three moles multiplied by the sum or sorry, multiplied by the bond entropy of a carbon hydrogen single bond added to this. We have one mole multiplied by the bond entropy of a carbon carbon single bond added to this. We have one mole multiplied by the bond entropy of a carbon oxygen double bond. And then added to this, we have one mole multiplied by the bond entropy of a carbon oxygen single bond. And then added to this, we have three moles multiplied by the bond entropy of an oxygen hydrogen single bond. So now we're just going to plug in our values just as we did before. So that in our next line, we would have five moles multiplied by the bond entropy of a carbon hydrogen single bond given us 413 kilojoules per mole, which is added to one mole multiplied by the bond entropy of a carbon carbon single bond given us 347 kilojoules per mole added to this. We have one mole multiplied by the bond entropy of a carbon oxygen single bond giving us 358 kilojoules per mole multi or added to this, we have one mole multiplied by the bond energy of a oxygen hydrogen single bond given as 467 kilojoules per mole. And then added to this, we have one mole multiplied by the bond entropy of an oxygen oxygen double bond given as 498 kilojoules per mole closing off our brackets and completing the sum of bond entropy of our reactants. We would now subtract from the sum of bond entropy of our products where we begin with three moles multiplied by the bond entropy of a carbon hydrogen single bond given in the prompt as 413 kilojoules per mole added to this, we have our one mole multiplied by the bond entropy of a carbon carbon single bond given to us as 498 um correction as 47 kilojoules per mole given from the prompt then added to this, we have one mole multiplied by the bond entropy of our carbon oxygen double bond given us 745 kilojoules per mole. And then added to this, we have our one mole multiplied by the bond entropy of a carbon oxygen single bond given us 358 kilojoules per mole. And added to this, we have our three moles multiplied by the bond entropy of an oxygen hydrogen single bond given us 467 kilojoules per mole closing off our brackets and completing the sum of bond entropy of our products. In our next line, we would simplify so that we have the result of the bond entropy of our reactants being kilojoules after we cancel out our units of moles, and then this is subtracted from the result of our second bracket being the sum upon entropy of our products as kilojoules, once we cancel out our unit of moles again. And so from this difference in our next line, we would simplify so that we have the results being negative 355 kilojoules of heat released per mole of our reaction. Again, this is the entropy change of our reaction in kilojoules per mole. And would be our second final answer. And so our two final answers are first being our entropy change of negative 85 kg per mole. And then our entropy change being equal to negative 355 kilojoules per mole as well. For the conversion of ethanol to acetic acid would be our two final answers to complete this example. I hope this made sense and let us know if you have any questions.

Struggling with GOB Chemistry?

Watch next

Bond Energy practice set