Here it says to calculate the change in the standard entropy of our reaction for the following reaction in 25°C. Here we have two moles of nitrogen monoxide reacting with one mole of oxygen gas to produce 2 moles of nitrogen dioxide gas. Here we're told that the enthalpy of reaction is equal to -114.14 kilojoules.
All right, so here they want us to calculate this. And remember, change in the standard entropy of our reaction is just equal to products, minus reactants. Here we're given the standard molar entropies of each of the compounds found within this reaction. We look at our products. We're going to say we have two moles of nitrogen dioxide, and according to our chart, each one has a standard MOA entropy of 240.1 joules over moles times K.
Plug that in, moles cancel out. Minus my reactants. We have two moles of nitrogen monoxide each one is 2108 joules over moles times K. Again, moles cancel out. Plus we have one mole of oxygen gas with a value of 205.2 Joules over moles times K. So again, moles cancel out. Here we can see that the moles cancel out, so our units at the end will be in joules per Kelvin.
When we plug this in, we get 146.6 Joules over Kelvin. This will be our final answer. Now, you might also notice that I gave us the enthalpy of the reaction. Here we don't even need to use it because the formula to calculate the entropy of our system doesn't need this entropy value. So again, sometimes professors give you additional information and that doesn't mean you have to use it. In this case, I gave you enthalpy of the reaction and it's just important. Is it just telling us at which the reaction is releasing energy? But it's not important to answer the question here.