Question

given the following reactions,

Answer 1

This is a Hess' Law problem. We need to manipulate the equations in such a way that all the compounds in them cancel out except for the compounds in the reaction of interest.

In this case, our reaction of interest has N2O on the left side. The only equation with N2O is the first one, so we need to flip it (products and reactants switch sides). This also means we need to switch the sign of ΔG, making it +139.56 kj. Here is what the first equation looks like now:

\[2N_{2}O+O_2 \rightarrow4NO \space \space \space \space \space \space \space \space \space \space \Delta G=139.56 \space kJ\]
Now, looking again at our reaction of interest, you can see that NO2 is on the right side. The only reaction with NO2 is the second one, and it already has NO2 on the right side so we don't need to flip it. Both the first and second equation have NO, but NO is not in our equation of interest - this means it must cancel out somehow. There are 4 NO in the first equation, but only 2 in the second. We can multiply the second equation by 2 to get 4 NO, which will cancel out the 4 NO in the first equation. This also means we need to multiply ΔG by 2, giving -139.4 kJ. Our second equation looks like this now:

\[4NO+2O_2 \rightarrow 4NO_2 \space \space \space \space \space \space \space \space \space \space \Delta G=-139.4 \space kJ\]
Finally, we can add these two reactions to get our reaction of interest! The 2N2O stays on the left side, the O2's from each equation give 3O2 total (they're added together since they're both on the left side), the 4NO cancels (since it is on opposite sides of the equation), and 4NO2 stays on the right side. To get ΔG for the reaction of interest, we add the ΔG's from each of our other reactions: ΔG = 139.56 kJ - 139.4 kJ = 0.16 kJ.

The answer is B. If anything is unclear please let me know! Hope that helps!

Answer 2

thank you soo much :)