Methanol (CH₃OH) is made industrially in two steps from CO and H₂. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane (CH₄):

Step 1. CO(g) + 2 H₂(g) S CH₃OH(l) ΔS° = - 332 J/K

Step 2. CH₃OH₁l₂ → CH₄(g) + 1/2 O₂(g) ΔS° = 162 J/K

(k) Calculate an overall ΔG°, ΔH°, and ΔS° for the formation of CH₄ from CO and H₂.

Video transcript

Hello. In this problem we are told the industrial preparation of acetic acid is a two step process. It is the chief component of vinegar. Were asked to calculate the values for the change in entropy entropy and gibbs free energy. Under standard conditions for the overall reaction, we can find the overall reaction by looking at two step process and the first step we see that methanol is produced and then the second step it is consumed. So that will cancel. If we combine everything that's left over, we get the overall reaction. So we have two moles of carbon monoxide gas reacting with two moles of hydrogen gas, give us one mole of acetic acid. And so since the overall reaction is the sum of these two steps, that means then that our change in entropy for the reaction is equal to the sum of the change in entropy for step one plus that for step two. So the change in entropy then for reaction is equal to negative two or 19.0 joules per kelvin plus a negative two and 77.4 joules per kelvin. That then works out to -496.4 Joules per Kelvin. So this is then our change in entropy for the overall reaction. Next then we're gonna find the change in the empathy for the reaction Under standard conditions that is equal to then some of the change in entropy of formation for products minus the sum of the change in entropy of formation for our reactant and so make use of the values that are provided. So we have one mole of our product acetic acid times its value of negative 484. modules per mole most attractive. Our reactant. We have two moles of Carbon monoxide times this value of negative 110.5 kamal. And then we'll add to that our other reactant. So we have two moles of hydrogen gas times its value which is zero. And so our units of moles will cancel. We'll find then that this works out to negative 263.5 no jules. So this is in our change in entropy. Under standard conditions for the reaction overall reaction. And then we will do similarly to find the change in get free energy for the reaction. As we do with entropy. We add up the change and good for energy for our information for our products and subtract off. Then change and give free energy information for our reactant. So making use of the values that are provided. We have then one mole of acetic acid times its value. Have a negative here in 90 Phil jules primal, we set out to attract off our reactant. We have two moles Carbon monoxide times its value. 137.2 jules from all. And then we add to that two moles of hydrogen and its value is zero Phil jules from all our units of moles cancel. Good luck with units and kill jules. This works out to negative 115.6 pill jules. So this then, is our value for the change in gibbs. Free energy for the reaction. Under standard conditions. Thanks for watching. Hope. This helps.