In photosynthesis, plants form glucose (C6H12O6) and oxygen from carbon dioxide and water. Write a balanced equation for photosynthesis and calculate ΔH°rxn, ΔS°rxn, and ΔG°rxn at 25 °C. Is photosynthesis spontaneous?

Video transcript

Hi, everybody. Welcome back. Our next problem says in photosynthesis, assume that plants form sucrose C 12 H 2211. And oxygen from carbon dioxide and water write a balanced equation and calculate delta H standard of the reaction. Delta S standard of the reaction and delta G standard of the reaction at 25 °C is this reaction spontaneous. We have four multiple choice options with different values. For all three of these things. We're asked to calculate and different answers as to whether the reaction is spontaneous. We'll look at that in more detail as we work through our problem. Our first step is a balanced equation. So our reactants are carbon dioxide and water. So, all right, CO2 plus and let's see the gap for the coefficients we need to put in plus water goes to sucrose C 12 H 22 0 11. Oxygen don't forget. Oxygen is always comes in that diatomic form 02. You start by balancing carbon on the left side, we have one carbon from CO2 right side, 12 carbons from sucrose. We can just give co two a coefficient of 12. And next, we'll balance our hydrogens. The left side, water gives us two hydrogens, sucrose gives us 22. So we can look at just adding a coefficient of 11 to water to give us 22 on both sides. And then if we look at our oxygens last, it's a little more complicated since there's oxygen in all our products and reactants. For CO2, we have 12 times 2, 24 oxygens, the water, we have 11 times 1, 11 sucres gives us 11 and our 02 gives us two. So our elevens balance each other out. We have 24 and two. So if we give 02, a coefficient of 12 will balance our oxygens out. So not too bad there. So we have the first part of our answer. Our balanced equation 12 co two plus 11, H2O goes to C 12 H 22 0 11 plus 1202. So now I can move on to calculating my first value delta H the entropy of reaction. And I get to that by subtracting the delta H of my reactants, the delta of formation of reactants from the delta H formation of my products. So delta H formation of my products minus delta formation of my reactants. It's important to remember that I have to multiply these values by the coefficients, my balanced equation, the delta H of my reaction will equal first, we do the delta H formation of my products brackets there. These are values that I can look up for each of my products and reactants. So I start with sucrose, I have one mole multi four, there's no coefficient. So coefficient is essentially one. My delta H formation of sucrose is negative 2226 0.1 kilojoules per mole. Now, then I add to it the delta a formation of my oxygen but oxygen is in the form 02, it's that pure elemental form and its standard state. So it has a zero anal pia formation. So plus zero there close brackets, then I subtract that value for my reactants. So open brackets. Now for a CO2, I have 12 moles, my coefficient of 12 multiplied by that value for CO2, just negative 393.5 kilojoules per more plus. Now the value for water 11 moles multiplied by negative 285.8 kilojoules per mole. So now we'll just simplify these terms a bit and that gives me negative 2226.1 kilojoules. Note that my moles have canceled out minus and then in parentheses, negative 7865.8 kilojoules, keep track of those negative signs there because I have I subtracting a negative. So that turns into adding and my final answer here is 5639.7 kilojoules. So that is my entropy DH of the reaction. So that's my first value. So we can go ahead and compare that to our multiple choice options. Choice A for this value has negative 10,091 0.9 kilojoules. So that's incorrect. And we can eliminate. Choice. A choice B has negative 5639.7 kilojoules. Not correct because our value is positive as the wrong sign. So choice be incorrect. Choice C delta H of their action is negative 10,091 0.9 kilojoules. So that's incorrect. Cross out choice C. And just by process of elimination, we know that choice D must be our answer. See it as delta H of the reaction positive 5639.7 kilojoules. So if I run a test, I'm done at this point, I have my just by elimination, my answer is choice D. But since this is a practice problem, we'll keep going and work our way through the rest of the values here. So the next value I'm looking for is delta S of a reaction entropy change, scroll up a bit to make room there. So delta S standard of our reaction is similar, it's equal to the entropy. So not delta S but just S standard of my products minus the entropy of my reactants. And again, remember to multiply by the coefficients that will equal one more multiplied by we're looking at sucrose here. Again, these are values we look up for these chemicals. So the entropy of sucrose is 360 0.24 and the units here are joules per mole. Kelvin plus now for oxygen and unlike the entropy of formation oxygen has a value for entropy. So 12 moles or excuse me, 11 moles, no, sorry, 12 moles getting turned around. We gotta keep focusing on the fact that this is products first 12 moles of 02 and its S value is 205.2 jules, primo Kelvin brackets around that minus. Now, let's look at the values for reactants, 12 moles multiplied by the value for co two, 213.8 jewels. Primo Kelvin plus the value for water. 11 wolves multiplied by 20.0 jewels for more. Kelvin. Let's simplify these terms here. Notice the moles have canceled out. I have 2822.64 joules per Kelvin minus 3335.6 joules per Kelvin, which gives us a final value of negative. Notice, my second term is bigger 512 0.96 Jules per Kelvin. So that's going to be my answer for delta S. However, notice that when I get to delta G, I'm going to need both my delta H and delta S values and they're in different units. Delta H has kilojoules, delta S has joules. And if I look at the units of my delta G, it's kilojoules. So as long as I'm here, I'm just going to do a quick conversion factor to convert this to kilojoules. So I will multiply this by one kilojoule divided by 1000 joules, scroll up a little bit there and that will give me a value of 0.51296 kilojoules per kin that I'll be using in my next equation. Let's double check. Does this match my answer in choice? D yes, negative 512.96 Joules per Kelvin. So finally on to my last quantity here, delta G standard of my reaction, my gibbs free energy which is equal to my change in enthalpy delta H standard of my reaction minus T delta S of the reaction. The last value I need to kind of do something to is my temperature, my temperature and my reaction if I scroll up was 25 °C. But I can see from my other values that I need that express name Kelvin. So let's just do a quick conversion here. 25 °C plus 273.15 equals 298.15 Kelvin. So that's the temperature I'll use there. So my delta G I have my delta H of the reaction 5639.7 kilojoules minus my temperature is 298.15. Kelvin multiplied by my delta s of the reaction which again, I converted kilojoules, 0.151296 kilojoules per Kelvin check my units kill the tools in the first term, my Kelvins cancel out and I have killer jewels in the second term so that I can subtract them. And that gives me a final answer of 5000 792.6 kilojoules. So when I look at choice D that does match what's here. And finally, let's note that my delta G of the reaction is positive, which means that it is non spontaneous heat must be, or energy must be put into the reaction to make it go forward. And of course, that makes sense because we're talking about plants generating sucrose, sucrose is a molecule that stores energy. So energy in the form of light needs to be put into this reaction to make it take place. So once again, my answer here is choice D calculated my delta H my delta S my delta G and discovered that this reaction is non spontaneous as we combine carbon dioxide and water to make sucrose and oxygen. See you in the next video.