Methanol 1CH3OH2 is made industrially in two steps from CO and H2...

Question

Methanol 1CH3OH2 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 1CH42: Step 1. CO1g2 + 2 H21g2 S CH3OH1l2 ΔS° = - 332 J>K Step 2. CH3OH1l2 S CH41g2 + 1>2 O21g2 ΔS° = 162 J>K (l) Is the overall reaction spontaneous at 298 K?

Accepted Answer

Welcome back everyone sulfuric acid, H two S. 04 can be produced via the contact process. For the first step, sulfate is made by burning sulfur. We have a three step mechanism given and the associated entropy values for each reaction determine whether the overall reaction is spontaneous at 2 98 kelvin and notice that we're given our entropy formation values for each of our re agents in each step of the mechanism. Let's recall that finding gibbs free energy change for reaction will tell us about the spontaneity of a reaction where specifically having a negative gibbs free energy change. So this is a negative sign here tells us that our reaction is spontaneous. Next recall that if we have a gibbs free energy change equal to zero, our reaction is at equilibrium. And if we have a gibbs free energy change that is positive or greater than zero than our reaction is going to be non spontaneous. And so in order to find gibbs free energy change, we want to recall our gibbs free energy change formula where we find delta G. And it's set equal to our entropy change delta H. Of our reaction minus the temperature in kelvin times the entropy change delta S. So we need to begin by finding our entropy change for each step. Beginning with step one, recall that we can find its entropy change by taking the sum of our and to be a formation of our products subtracted from the sum of the entropy of formation of our reactant. And so following step one we have delta H equal to beginning with the entropy of formation of our only product being S. 02 gas. We see that we have just one mole of it in step one. So we have one mole multiplied by the entropy of formation of sulfur dioxide given in the prompt as negative 2 96.8 units of kilo jewels per mole. So this completes the some of the entropy of formation of our products. And now we want to subtract by the sum of the entropy of formation of our reactant. We have to react ints beginning with our solid sulfur. We have just one mole of that. And actually just for clarity we should write our product next to the moles. So we'll add in one mole of S. 02 for our only product. And then for our first reactant we have one mole of sulfur. This is multiplied by the entropy of formation of solid sulfur Given in the prompt zero killer joules per mole. And then we're going to have to make more room. So added to this. And let's put in our bracket, we have our second reactive oxygen gas. We have one mole of oxygen gas, multiplied by its given anthem P. A. Formation also as zero killed jules, per mole. And this completes to some of our entropy of formation of our reactant. And so we would find that our entropy change for step one of the reaction is equal to negative 296.8. And notice that our units of moles cancel out. So we're left with units of killing jewels. For our entropy change now we can go and calculate our gibbs. Free energy change for step one which is set equal to our calculated entropy above as negative 2 96.8 kg, joules subtracted from our temperature in kelvin. Were given a temperature in kelvin of 2 98 kelvin and then multiplied by the entropy change. For step one being 10 positive 10 simple plug that an S 10.9 with units of joules per kelvin. Now we want to recall that our unit for gibbs free energy changes in killing jewels. And so notice that because we have jewels for our entropy change we need to convert to kill a jewels that are units of jewels can cancel. And so we want to multiply by a conversion factor to go from jewels in the denominator to kill the jewels in the numerator for our entropy change. And so we would recall that our prefix kilo tells us that we have an equivalence of 10 to the third power of our base unit jewel. And so canceling out jewels were left with killer joules per kelvin for our entropy change now we're going to cancel out our kelvin units. And then we're going to take the products of our terms being multiplied first. And so in our next line we would have our gibbs free energy change for step one equal to negative 96.8 kg subtracted from a value of 3.2482 would be the result of our products that we underlined and next week just will take the difference between these two terms and that's going to give us our Gibbs free energy change and we'll write it below Equal to a value equal to negative 300.0482. And we have units of illegals. So we're going to need to calculate more to more gibbs Free energy change values for the rest of our steps before we can come up with an overall reaction and an overall gibbs free energy change for that reaction to lead us to our final answer. So let's move on to step two now. So for step two we need to find the entropy value. Let's separate our work. So we begin with our entropy change equal to beginning with our some of our entropy change of our products. We have two moles according to the pumped of Sulfate gas, multiplied by the entropy change of sulfate gas given in our prompt as negative 3 95.7 kg jules Permal. This is our only product and this would complete the some of the entropy change of products. And sorry, just to make a correction I just noticed our formula. Our parentheses should be here for entropy change. Okay, so subtracting now from the entropy change the some of the entropy change of our reactant. We begin with our first reactant. First step to being two moles of sulfur dioxide gas, multiplied by the entropy of formation for sulfur dioxide given in the prompt as negative 2 96.8 kg joules per mole. And just going to make more room adding to this, we have our second reactant in which for step two we have one mole of oxygen gas, multiplied by the entropy of formation for oxygen gas given as again, zero kg per mole. This completes the sum of entropy change of our reactant. And now we're going to just Plug this into our calculators in which for our first bracket we get a value of -791.4. And we have two students of killer jewels after we cancel out moles. And we don't need brackets anymore. And then this is subtracted from our some of our entropy change of our reactant in which we get a value of negative 593.6 kg after we cancel out most. And so now taking the difference between these values will get our entropy change for step two equal to the value negative 197.8 kg joules. Now we're going to go right into calculating our gibbs free energy change for step two. So recall that that's delta G equal to our calculated entropy change from above one negative, 197.8 kg for step two subtracted from our temperature to 98 kelvin. And then we want to multiply by our given entropy change for step two which above was given as negative 188 Jules per Kelvin. Again we need to multiply by the conversion factor to cancel out jewels and the denominator and kill jewels in the numerator. So again, one killer jewel is equivalent to 10 to the third power of our base unit jewel. Getting rid of jewels and kelvin were left with kill a jewel units. And we're just going to begin by taking the product of our parentheses terms. So we'll have negative 197.8 kill jewels on the right hand side, subtracted from the product of our parentheses, which results in a value of negative 56. kg joules. Now taking the difference between our terms in the next line, we would get a gibbs free energy change for step two equal to negative 141.776 killed jules. So now we want to follow suit for step three. So we'll do that below here, Step three, calculating the entropy change for step three, beginning with the sum of our entropy change of our products or entropy of formation of our products. We have our first product for step three or our only product for step three being for sulfuric acid, which we have above. Just one mole of. So one mole of sulfuric acid. H two S. 04 multiplied by its entropy of formation being negative 814 point oh joules per mole. This completes a sum of entropy of formation of our products and now subtracting from the sum of entropy of formation of our first reactant for step three which is sulfate gas. So that's one mole of S. 03 gas multiplied by sulfates entropy information given in the prompt as negative 395.7 kg per mole. And now we want to add to this our entropy of formation of our second reactant for step three being liquid water. And I'm just going to make some more room. So we have our one mole of liquid water, one mole of H 20. Multiplied by its given entropy of formation in the prompt as negative 285.8 kg joules per mole. So this completes the sum of entropy of formation of our reactant. So for our first bracket we should come up with just negative 814.0 kg joules per mole And then minus our some of entropy information of our products, which results in a value of negative 681.5 And we just have units of kg. So apologies, we need to correct our some of entropy of formation of our products because it just should have units of kill jules after we cancel out our units of moles for both our products and react ints. So now we will just take the difference between our killer jewel values and we'll find our entropy change for step three equal to a value of -132.5. Kill it jules. And so now calculating gibbs free energy change for step three we have delta G equal to our calculated entropy change for step three being negative 132.5 kg jewels subtracted from our temperature given us 2 98 kelvin multiplied by our entropy change for step three in which were given a value of negative 170. So let's put that in a negative 170 joules per kelvin. And then multiplying by our conversion factor to go from jewels in the denominator to kill jewels in the numerator for one kg jule we have an equivalent of 10 to the third power jewels. So canceling out jewels and canceling out kelvin, we're left with killer jewels. And then beginning by taking that products for our first step we have negative 132.5 kg jewels on the right hand side subtracted from the results of negative 50.66 killer jewels. And then taking the difference we get our gibbs free energy change for step three equal to negative 81.84 killed jules. So now we're just going to scroll back to our given reaction, Steps one through three, and we want to come up with our overall reaction. But in order to do so, we need to cancel out what we can make. Similar in each step. Notice that in Step two we have two moles of sulfuric dioxide gas, but in sorry sulfur dioxide gas. But in Step one, we have just one mole of sulfur dioxide gas on the product side. So we're going to take the entirety of Step one and multiply it. And sorry, the entirety is step one here in parentheses. And we're going to multiply by a factor of two, which means we also must take our given entropy change for step one and multiply it by a factor of two. And that's going to give us two moles of solid sulfur plus two moles of oxygen gas for step one. And that will yield Two moles of sulfur dioxide gas on our product side. And just to make a correction, we will not multiply our entropy change by two. But rather our gives free energy calculation for step one. And so we calculated that to be negative 300.0482. So we're going to multiply by two. And that's going to give us our gibbs free energy change for step one now equal to negative 600.9 64 kg jewels. And so we can cancel out now are two moles of sulfur dioxide gas. With our two moles of sulfur dioxide gas from Step two on the reactant side. And now looking at step three, notice that we have just one mole of sulfur, or sorry, sulfate, S. 03 gas. And in Step two we have on the product side two moles of sulfate S. 03 gas. So we're going to need to also take step three and multiply everything by a factor of three, including the calculated gibbs. Free energy change that we found for step three. And that was below as negative 81.84 killed jules, which we multiplied by three. And we get a gibbs Free energy change equal to a value. And sorry correction, I'm meant to say we multiplied by two because notice that above we have two moles of sulfate gas. So we're multiplying by a factor of two. So apologies for that mistake When we multiply our gibbs. Free energy change by two. For step three, we get negative 163.68 kg joules now for step three. And now, riding out what Step three looks like after we multiply the reaction by a factor of two. We will now have two moles of sulfate gas plus two moles of liquid water, yields two moles of sulfuric acid, H two, S. 04. And so we can now cancel out also are two moles of sulfate from Step two, with our two moles of sulfate from Step three on the react inside. So now, with the exception of everything we canceled out, we can come up with our overall reaction in which we have all of our reactions added together. So two moles of solid sulfur plus our one mole of oxygen, gas, plus two moles of water. This is going to yield our products in which we have, the only product left being from Step three, which is our two moles of sulfuric acid H two, S. 04. So we're going to move all of our information below being our full reaction are full overall reaction and our calculated entropy gibbs free energy changes. Sorry. So we have here our overall reaction and now we need our overall gibbs free energy change for this reaction, in which we take the some of the gibbs free energy change calculated for each of our steps for step one, we found after multiplying by 2 600.964 kg joules four. Step two, we found negative 141.776 kg, which we did not manipulate. And then for step three we multiplied by a factor of two and got negative 163.68 kg joules. So now taking the sum of these three values, we get our overall gibbs free energy change equal to negative 905.6 illegals. And notice that our Gibbs free energy change is less than zero or in other words it's negative. And so therefore we can say that our reaction overall is going to be spontaneous. And so this would be our final answer to complete this example, that our reaction is going to be spontaneous at 298 Kelvin, and our answer corresponds with choice B in the multiple choice as the correct answer. So I hope that this made sense and let us know if you have any questions.

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