Please help me
N2O5(g) + H2O(l) --> 2HNO3(l) delta H° = -76.2 kJ
H2O(l) --> H2(g) + 1/2O2(g) delta H° = 286.0 kJ
1/2N2(g) + 3/2O2(g) + 1/2H2(g) --> HNO3(l) delta H° = -174.0 kJ
Calculate delta H° for the reaction
2N2O5(g) --> 2N2(g) + 5O2(g)
See the post above.
To calculate the delta H° for the reaction 2N2O5(g) --> 2N2(g) + 5O2(g), we can use the Hess's law, which states that the enthalpy change of a reaction is independent of the pathway taken, as long as the initial and final conditions are the same.
We can start by examining the given reactions and rearranging them to match the desired reaction:
1) N2O5(g) + H2O(l) --> 2HNO3(l) (Eq. 1)
2) H2O(l) --> H2(g) + 1/2O2(g) (Eq. 2)
3) 1/2N2(g) + 3/2O2(g) + 1/2H2(g) --> HNO3(l) (Eq. 3)
Now we need to manipulate these equations to obtain the desired equation:
Multiply Eq. 2 by 2 to balance the hydrogen and oxygen atoms:
4) 2H2O(l) --> 2H2(g) + O2(g)
Since we need 2N2(g) on the product side, multiply Eq. 3 by 2:
5) N2(g) + 3O2(g) + H2(g) --> 2HNO3(l)
Finally, add Eqs. 4 and 5 together to match the desired reaction:
6) 2H2O(l) + N2(g) + 3O2(g) + H2(g) --> 2HNO3(l) + 2H2(g) + O2(g)
Now, we can rearrange Eq. 6 to match the desired reaction format:
7) N2(g) + 2H2(g) + 3O2(g) --> 2HNO3(l) + 2H2O(l) + O2(g)
Now we can sum the equations and the enthalpy changes:
2 × Eq. 1 -> 2N2O5(g) + 2H2O(l) → 4HNO3(l)
-174.0 kJ + 2 × 286.0 kJ = 398.0 kJ
Eq. 7 -> N2(g) + 2H2(g) + 3O2(g) → 2HNO3(l) + 2H2O(l) + O2(g)
-76.2 kJ + 398.0 kJ = 321.8 kJ
Therefore, the delta H° for the reaction 2N2O5(g) → 2N2(g) + 5O2(g) is 321.8 kJ.