Solve the heat combustion equation: 2N2+5O2→2N2O5 ΔH=? N2O5+H2O→2HNO3 ΔH= -76.6KJ H2+1/2O2→H2O ΔH= -285.8KJ 1/2N2+3/2O2+1/2H2→HNO3 ΔH= -174.1KJ
Sorry but you didn't proof your post and the last equation didn't post correctly. I have copied it here. /2O2+1/2H2→HNO3 ΔH= -174.1KJ
My best guess is that you meant 1/2N2 + 3/2O2 + 1/2H2 = HNO3 with delta H of -174.1 kJ but I don't know that and I'm hesitant to do all that work for nothing. Please repost your program and proof it to know it is correct.
To solve the heat combustion equation, we need to use Hess's Law, which states that the enthalpy change of a reaction is independent of the pathway between the initial and final states. Hence, we can use the given reactions and their enthalpy changes to find the enthalpy change of the desired reaction.
First, let's list the given reactions along with their corresponding enthalpy changes:
1. 2N2 + 5O2 → 2N2O5 ΔH = ?
2. N2O5 + H2O → 2HNO3 ΔH = -76.6 kJ
3. H2 + 1/2O2 → H2O ΔH = -285.8 kJ
4. 1/2N2 + 3/2O2 + 1/2H2 → HNO3 ΔH = -174.1 kJ
Now, we need to manipulate the given reactions and their enthalpy changes to obtain the desired reaction:
1. Multiply reaction 2 by 2 to cancel out HNO3 and match the number of N2O5:
2N2O5 + 2H2O → 4HNO3 ΔH = -153.2 kJ
2. Multiply reaction 1 by 2 to cancel out N2O5:
4N2 + 10O2 → 4N2O5 ΔH = 2 * ?
3. Multiply reaction 3 by 2 to match the number of H2O:
2H2 + O2 → 2H2O ΔH = -571.6 kJ
4. Multiply reaction 4 by 4 to cancel out HNO3:
2N2 + 6O2 + 2H2 → 4HNO3 ΔH = -696.4 kJ
Now, add all the manipulated reactions and their enthalpy changes together:
4N2 + 10O2 + 2H2 + O2 + 2N2 + 6O2 + 2H2 -571.6 kJ -696.4 kJ -153.2 kJ = 0
Simplifying the equation gives: 6N2 + 14O2 = 1221.2 kJ
Therefore, the enthalpy change for the combustion of 2N2 + 5O2 to form 2N2O5 is ΔH = 1221.2 kJ.