Here it says to calculate the enthalpy of reaction for the overall reaction. So we have carbon monoxide plus nitrogen monoxide producing carbon dioxide and half a mole of nitrogen gas. Now here we're given these partial reactions below with their own known standard enthalpies of reaction. Now what do we do? Well, let's follow the rules.
Step one says that I need to start with the first compound in the overall equation and locate it in the set of partial reactions. Compound from partial reaction must match in terms of number and type, and we'll see what that means with the one from the overall equation. This may require you to reverse, multiply or divide the partial reaction, which will also affect your standard enthalpy value. Step two says keep moving on to the next compound in the overall equation until you locate all compounds in the partial reaction. So let's do steps one and two initially.
All right, so we have two. So if we go back to step one, it says that we need to locate the compound, we must locate the first component in the overall equation and locate it in the set of partial reactions. All right, so in my overall equation, I have carbon monoxide gas as a reactant, and there's one of it, 1 mole of it. So if I look in the partial reactions, we see carbon monoxide is right here. But we have an issue. Although there is one mole of it, like I want, it is a product, we need it to match up with the overall equation. We need it to be a reactant.
To do that, that means I'm going to have to reverse the reaction. Now, when I reverse the reaction, it's now going to become a reactant. O2 also becomes a reactant and CO2 becomes a product. Now remember, when I reverse the reaction, that's going to reverse the sign for delta H. So now my carbon monoxide, there's one mole of it. So the number is good, and the type is also good because it's also a reactant. Now, just like in my overall equation, this equation is gone now because it's been reversed and is represented by this bottom one.
Next, I need to find NO gas. I need it to be 1 mole of it and I need it to be a reactant. Here I see it. Two issues. First, it has a two in front of it. I need it to have a coefficient of one just like in my overall equation. Also, I need it to be a reactant. So what I'm going to do here is I'm going to divide by 2 and reverse. So here I'm going to divide by two and reverse it. So that's going to become NO gas. And then when I reverse and when I divide by two, this becomes half N2 plus half O2.
When I reverse the equation, remember that's going to change the sign, so this becomes negative. And when I divide by two, I have to divide delta H by two so this is -90.3 kilojoules. So now this NO matches up exactly with the overall equation. Keep going. We need one mole of CO2 as product which we have here. We need half a mole of N2 product which we have here. So all the compounds I need match up with the overall equation.
So now we go to step three, combine the partial reactions and cross out reaction intermediates if present. Reaction intermediates are compounds that look the same with one as a reactant and the other as a product. Once we do that, we go to step four and we add all the standard enthalpy values of my partial reactions to obtain the enthalpy of reaction of the overall reaction. Alright, so who are intermediates? Things that look the same, but one's a product and one's a reactant. We have half O2 here as a reactant, which will cancel out with this half O2. That's a product.
Those are our only reaction intermediates. Everything else comes down. If you look when everything comes down, it should give you back the overall equation you were looking for in the first place. It doesn't matter the order that you write them in. All that matters is that the coefficients match the overall equation and that they're on the right side. Reactants here should match up with the reactants up in the overall products here should match up with the products up here. All we do now is we add up these enthalpies of reaction. So when I add up -283 plus this -90.3 that gives me a total enthalpy of reaction equal to -373.3 kilojoules. So this would be our Hess's Law question, where we solved for the overall enthalpy of the reaction for the overall reaction.