Calculate the enthalpy change for the reactions

NO(g) + O(g) → NO2(g)
from the following data:
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = -198.9 kJ
O3(g) → 1.5O2(g) ΔH = -142.3 kJ
O2(g) → 2O(g) ΔH = 495.0 kJ

Put the answer in kJ.

-304.1kJ

Take equation 1 and add it to the reverse of equation 2 and add that to the reverse of 1/2 equation 3.

When reversing dH, change the sign.
When taking 1/2 equation, also take 1/2 dH.

Hmm, calculating enthalpy changes? That sounds like a job for a superhero, but you'll have to settle for me, Clown Bot! Let's break it down step by step:

1. We have the equation: NO(g) + O(g) → NO2(g).
2. We need to find the enthalpy change for this reaction, ΔH.
3. First, we notice that NO2(g) is on both sides of the equation, which means we can make use of the given equation: NO(g) + O3(g) → NO2(g) + O2(g), with a ΔH of -198.9 kJ.
4. We also notice that O2(g) is on both sides of this new equation, so we can use the equation: O2(g) → 2O(g), with a ΔH of 495.0 kJ.

Now, let's manipulate these equations to get the desired reaction:

First, we'll need to flip the second equation and multiply it by 2 to balance the number of O atoms:
2O(g) → O2(g) with a ΔH of -990.0 kJ. (Double the ΔH value)

Secondly, we'll need to combine the two equations:
NO(g) + O3(g) + 2O(g) → NO2(g) + O2(g) + 2O(g)

Lastly, we'll cancel out the common O2(g) and simplify the equation:
NO(g) + O3(g) + 2O(g) → NO2(g) + 3O(g)

Now, let's find the ΔH by adding the enthalpy changes:
ΔH = -198.9 kJ - 990.0 kJ = -1188.9 kJ

So, the enthalpy change for the given reaction is -1188.9 kJ. Ta-da!

To calculate the enthalpy change for the reaction you provided, we need to manipulate the given equations and their enthalpy changes to obtain the desired reaction.

First, let's write down the reaction we want to calculate the enthalpy change for:
NO(g) + O(g) → NO2(g)

Since we don't have the enthalpy change directly for this reaction, we need to combine the given equations to obtain it. Here's how you can do it step by step:

1. Reverse the first equation:
NO2(g) + O2(g) → NO(g) + O3(g)
This is done by changing the direction of the arrow and the sign of the enthalpy change. So, the new equation becomes:
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = 198.9 kJ

2. Multiply the second equation by 2:
2O3(g) → 3O2(g) ΔH = -2 * 142.3 kJ = -284.6 kJ

3. Multiply the third equation by 1.5:
1.5O2(g) → 3O(g) ΔH = 1.5 * 495.0 kJ = 742.5 kJ

4. Add the equations together to cancel out the intermediates:
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = 198.9 kJ
+ 2O3(g) → 3O2(g) ΔH = -284.6 kJ
+ 1.5O2(g) → 3O(g) ΔH = 742.5 kJ

The O2(g) term appears on both the left and right sides of the equations, so we can cancel it out:
NO(g) + O3(g) → NO2(g) + 2O(g) + 2O3(g) ΔH = 198.9 kJ - 284.6 kJ + 742.5 kJ

Simplifying the equation, we have:
NO(g) + O3(g) → NO2(g) + 2O(g) + 2O3(g) ΔH = 656.8 kJ

Now, we can compare this equation to the desired reaction:
NO(g) + O(g) → NO2(g)

We can see that the desired reaction is missing two oxygen atoms on the product side. Since the enthalpy change is proportional to the amount of substance reacting, we multiply the enthalpy change by a factor of 2 to account for the two missing oxygen atoms:
2 * 656.8 kJ = 1313.6 kJ

Therefore, the calculated enthalpy change for the reaction NO(g) + O(g) → NO2(g) is -1313.6 kJ.

NO+O3=NO2+O2 -198.9 kJ

O3=1.5 O2 -142.3 kJ
O2=2O 495.0 kJ
NO+O=NO2 ?

NO+O3=NO2+O2 -198.9 kJ
O3=1.5 O2 -142.3 kJ
2O=O2 -495.0 kJ
NO+O=NO2 ?

NO+O3=NO2+O2 -198.9 kJ
O3=1.5 O2 -142.3 kJ
1/2(2O=O2) 1/2(-495.0 kJ)
NO+O=NO2 ?

NO+ O̶3̶ = NO2+ ̶O̶2̶ -198.9 kJ
O̶3̶ =1̶.̶5̶O̶2̶ -142.3 kJ
O= ̶1̶/̶2̶ ̶O̶2̶ -247.5 kJ
NO+O=NO2 =-588.7 kJ

530kj