Methanol (CH3OH) burns in oxygen to form carbon dioxide and water. Write a balanced equation for the combustion of liquid methanol and calculate ΔH° rxn, ΔS°rxn, and ΔG°rxn at 25 °C. Is the combustion of methanol spontaneous?

Video transcript

All right. Hi, everyone. So this question says that methanol that C 30 burns in oxygen to form carbon dioxide and water. Write a balance equation for the combustion of liquid methanol and calculate delta H standard for the reaction delta S standard for the reaction and delta G standard for the reaction at 25 °C is the combustion of methanol spontaneous. All right. So first recall that we can calculate delta G standard. That's the jibs free energy by using both the enthalpy, that's delta H standard and the entropy delta S standard as well as the temperature. This formula says that delta G standard is equal to delta H standard subtracted by the temperature multiplied by delta as standard. But before we talk about calculating these values individually, let's go ahead and write out the chemical equation and balance it starting off with methanol C 30 as a liquid reacting with oxygen gas to produce carbon dioxide gas and water vapor. So when discussing how to balance a chemical equation, the idea is to keep count of each type of atom, right. In this case, we have carbon hydrogen and oxygen in this chemical equation. So the idea is to make sure that the amount of each atom is the same on both sides of the equation starting off with carbon. For example, we can see that there is one carbon atom on each side of the equation. Hi Eugene, on the other hand has four atoms of it on the left side and only two on the right side, whereas oxygen has three atoms of it on both sides. So at this point, the only element that is not balanced is hydrogen. So to fix this, I'm going to place a coefficient of two in front of H two because this fixes the balance of hydrogen, which means that I now have four on each side of the equation referring to hydrogen apps. But by changing the coefficient in front of water, I've also changed the amount of oxygen on the right side instead of three, we now have four. Now because the ratio between oxygen atoms on both sides of the equation is 3 to 4. I'm going to try doubling all of the coefficients in my equation to see if I can reach a greater common value. So that's two methanol. There are two moles of methanol, two moles of oxygen gas, two of carbon dioxide and four of water. So lets see how this affects our counts for each atom with respect to carbon. There are two carbon atoms on each side of the equation hydrogen or rather there are eight hydrogen atoms on the left side and eight on the right side. And for oxygen, we have six on the left side and eight on the right side. However, because the difference is only two, I can fix this by adding one more mole of oxygen gas, bringing my coefficient up to three because this add two more oxygens or two more oxygen atoms on the left side, meaning I have eight on both sides. So now we have the balanced chemical equation which we can now use to calculate Delta H standard delta S standard. And therefore Delta G standard recall that Delta H standard for the reaction can be calculated by taking the entropies of all products and subtracting that by the entropies of all reactants. The same is true for the entropy delta S standard for the reaction can be found by adding the entropies of our products and subtracting that from the total entropies of our reactants. So first, let's start with Delta H standard. But to do this, we're going to recall or have to recall some of the standard entropies of formation found from the literature, namely those of our reactants and products. All right. So here on the left side of the screen, I've written down the relevant delta H standard of formations or our reactants and our products in units of kilo joules per mole. So now on the right side of the screen, I'm going to do the same thing or a similar thing and write down the standard entropies or our reactants and our products. All right. So now on the right side of the screen are the standard entropy values for all reactants and products in units of jewels per mole Kelvin. So first, let's start by calculating delta H standard of the reaction. Now, as discussed previously to find the delta H standard of the reaction, we have to add together the standard entropies of formation were all products and reactants and take the difference between those quantities. So first goes the sum of the entropies for the products. Now, it's worth mentioning that the entropies must also be multiplied by their stoichiometry coefficients. So in the case of carbon dioxide, for example, I would take its delta H standard of formation which is negative 238.6 kilo joules per more and multiply this by its coefficient of two or to be a bit more specific, we can say two moles because notice how this makes units of moles cancel out. So we'll do the same thing for water vapor that's four moles multiplied by the delta H standard of formation of water. But very quickly, I want to go back because I realized that I used the wrong delta H standard of formation for carbon dioxide. It's actually negative 393.5 kilojoule per mole. But going back to water vapor that is negative 241.8 kilojoules per mole. So next is the sum of the standard entropies of formation for our reactants, that's methanol and oxygen. So for methanol, that's a coefficient of two moles multiplied by its delta H standard of formation, which is actually negative 238.6 kilojoules per mole. And then for oxygen gas, we have three moles multiplied by zero kilo jules perm units of moles cancel out throughout. So after finding each individual sum and then evaluating the difference. Delta H standard of the reaction is equal to negative 1277 kg jules. So now let's do the same thing for delta S standard of the reaction starting off with our products. That's two moles for carbon dioxide. And the standard entropy of carbon dioxide is 213.8 joules per mole. Kelvin or add this to four moles of water multiplied by the standard entropy of water, which is 188.8 jules per mole. Kelvin. So now the sum of our product entropies gets subtracted by the sum of our reactant entropies. So for methanol, we have two moles of methanol multiplied by its standard entropy which is 126.8 joules per mole. Kelvin added to three moles of oxygen multiplied by its standard entropy of 205.2 Jules per mole. Kelvin notice how units of moles cancel out. So after evaluating each individual sum and taking the difference between them delta as standard of the reaction is equal to 313.6 jewels per Kelvin. However, to keep units consistent, let's convert this quantity from joules per Kelvin into kilojoules per Kelvin. Now recall that to, to do this, we're going to have to divide by 1000, but that is equivalent to taking our decimal place and moving it three spaces to the left. So this equals 0.3136 kg joules per Kelvin. So once again, on the topic of keeping our units consistent, we're going to have to convert our temperature that was 25 °C into Kelvin. So T is equal to 25 °C added to 273.15 which equals 298.15. Kelvin or simply 298 Kelvin. So last but not least delta G standard for the reaction is equal to delta H standard of the reaction subtracted by T multiplied by delta S standard of the reaction. So delta H standard of the reaction is negative 1277 kilojoules subtracted by T thats 298 Kelvin multiplied by delta S standard of the reaction which is 0.3136 kilo jewels per Kelvin. And lastly, after evaluating this expression, delta G standard of the reaction is equal to negative 1370 kg joules. Now recall that when Delta G standard for the reaction is negative. In other words, it's less than one. This means that the reaction is spontaneous and there you have it. So our final answer is as follows. Delta H Standard of the reaction is equal to negative 1277 kg joules. Delta S standard standard of the reaction is equal to 313.6 joules per Kelvin or 0.3136 kg jewel per Kelvin and Delta G standard. Standard of the reaction is equal to negative 1370 kg joules, which indicates that it is spontaneous the reaction. So if you stuck around until the end of this video, thank you so very much for watching and I hope you found this helpful.