Please help me! I don't know how to do this at all.
Calculate the enthalpy for the reaction:
NO (g) + O (g) → NO2
Given the following known reactions:
NO(g)+O3 (g)→NO2 (g)+O2 (g) 03 (g)→1.5O2 (g) O2 (g)→2O(g)
ΔH°=-198.9kJ
ΔH°=-142.3kJ
ΔH°=495.0kJ
It would help if you wrote questions that could be interpreted. You might try looking at the post to see how it is shown. What you have makes no sense; I have hit the enter button to try and separate your one reaction into three separate reactions. I hope I hit the enter key at the right place.
NO(g)+O3(g)→NO2(g)+O2(g) dH = -198.9 kJ
03(g)→1.5O2(g) dH = -142.3 kJ
O2(g)→2O(g) dH = 495.0 kJ
Add equation 1 to the reverse of equation 2 to the reverse of 1/2 of equation 3.
If you multiply the equation remember to multiply the dH value by the same number. If you reverse an equation change the sign of dH.
Add all of the new dH values to find the overall dH for the desired reaction.
To calculate the enthalpy for the reaction NO (g) + O (g) → NO2, you can use the concept of Hess's Law. This law states that the enthalpy change of a reaction is independent of the pathway taken.
Here's how you can use the known reactions and their enthalpy changes to find the enthalpy for the given reaction:
1. Start by writing the given reaction and its known reactions:
Reaction: NO (g) + O (g) → NO2 (g)
Known Reactions:
a) NO (g) + O3 (g) → NO2 (g) + O2 (g) (ΔH° = -198.9 kJ)
b) O3 (g) → 1.5 O2 (g) (ΔH° = -142.3 kJ)
c) O2 (g) → 2 O (g) (ΔH° = 495.0 kJ)
2. Manipulate the known reactions so that their stoichiometry matches the target reaction:
Multiply reaction a) by 2, and reaction b) by 2:
2NO (g) + 2O3 (g) → 2NO2 (g) + 2O2 (g) (ΔH° = -397.8 kJ)
2O3 (g) → 3O2 (g) (ΔH° = -284.6 kJ)
3. Combine the manipulated reactions to cancel out common compounds:
Add the manipulated reactions to get the desired reaction:
2NO (g) + 2O (g) + 2O3 (g) → 2NO2 (g) + 3O2 (g) + 2O2 (g)
2NO (g) + 2O3 (g) → 2NO2 (g) + 5O2 (g) (ΔH° = -397.8 kJ - 284.6 kJ)
4. Simplify the equation:
Combine the terms with the same compound on both sides:
2NO (g) + 4O3 (g) → 2NO2 (g) + 5O2 (g) (ΔH° = -682.4 kJ)
The enthalpy change for the reaction NO (g) + O (g) → NO2 is -682.4 kJ.