Use standard enthalpies of formation to calculate ΔH°_rxn for each reaction. b. SO₂(g) + 1/2 O₂(g) → SO₃(g)

Video transcript

Hi everyone for this problem, we're told that methanol is produced globally by reacting a mixture of carbon monoxide and hydrogen over a transition metal catalyst provide a balanced chemical equation for the combustion of methanol and determine the value of the standard entropy change for the reaction. And so for our balanced chemical equation, we're going to have two moles of ethanol, Which is CH three O H Plus three moles of oxygen gas is going to yield two moles of carbon dioxide plus formals of liquid water. Okay. And in order for us to calculate the standard entropy change for the reaction, which is our delta H of our reaction, it's going to be the sum of our products minus the sum of our reactant. And I'll explain what that means. So we're going to take the molds of our products and react ints and multiply them by their standard heats of formation and these are values that we are going to have to look up. So our standard heat of formation for carbon dioxide is negative 393.5 kila jules Permal. Our standard heat of formation of liquid water is negative 285. killer jules Permal. In our standard heat of formation of methanol is Going to be negative, 200 and 38 0. killer jewels per And anything that we have in its elemental state such as our 02 gas is going to be zero. So now we have all of our standard heats of formation. Now we can go ahead and solve for a standard and therapy change. So our standard entropy change Is going to be the sum of our products. So our first product is liquid water and we have four moles of it. So four moles of liquid water multiplied by its standard heat of formation, negative 285.8 plus. Our second product is carbon dioxide and we have two moles of it. And it's standard heat of formation is negative 393.5 kg joules per mole. So that's the sum of our products minus the sum of our reactant. So we have three moles of 02 gas and we said anything in its elemental state is going to be zero for its standard heat of formation. And then we have two moles of methanol and its standard heat of formation is negative. 238.4 kg joules per mole. So this is the sum of our products minus the sum of our reactant. And once we do this calculation we get a standard entropy change of negative 453 0.4 kila jules per mm. This is our final answer and that is the end of this problem. I hope this was helpful