Calculate ΔHrxn for the reaction: Fe2O3(s) + 3 CO( g)¡2 Fe(s) + 3 CO2( g) Use the following reactions and given ΔH's: 2 Fe(s) + 32 O2( g)¡Fe2O3(s) ΔH = -824.2 kJ CO( g) + 12 O2( g)¡CO2( g) ΔH = -282.7 kJ

Video transcript

Hello, everyone today with the following problem, calculate the change in entropy for the following reaction, use the following reactions and given entropy changes. So we first have to look at our reactions and compare them with the overall reaction. So with reaction one, we see here that we have our iron, our ferrous oxide on the product side. So our over our in our overall reaction, we have it on the reactant side. So we need to essentially reverse the sign of entropy such that it is positive 8 24.2 kilo joules. Looking at our second reaction, we look at the reactant side and the overall reaction has three moles of carbon dioxide or overall reaction. Well, our reaction two has only one more. And so we need to multiply the second reaction. And the envoy by three recall that we have to multiply everything by the same variable. So that would essentially give us a change in entropy of negative 8 48.1 kilojoules. And so we also have to recall hess's law that states that the entropy change of the net reaction is the sum of the entropy of the steps. So if you want to find the ultimate or the final entropy of the reaction, we take our reaction one which is 824 0.2 killer jules. And we add that to the second reactions entropy which is negative 8 48.1 kilojoules. And in doing so, we arrive in an answer of negative 23.9 kilojoules or answer twice a overall, I hope it's helped. And until next time.