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)
Dr. Bob I need your help please
PLEASE
DR BOB I NEED YOU
Calculate ∆H0
for the reaction
2 N2(g) + 5 O2(g) −→ 2 N2O5(g)
given the data
H2(g) + 1
2
O2(g) −→ H2O(ℓ)
∆H0
f = −285.6 kJ/mol
N2O5(g) + H2O(ℓ) −→ 2 HNO3(ℓ)
∆H0 = −75.9 kJ/mol
1
2
N2(g) + 3
2
O2(g) + 1
2
H2(g) −→ HNO3(ℓ)
∆H0
f = −173.4 kJ/mol
Answer in units of kJ.
To calculate the delta H° for the reaction 2N2O5(g) -> 2N2(g) + 5O2(g), you can use the Hess's Law which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps.
Step 1: Start with the given equation and the known enthalpy changes of the given reactions:
N2O5(g) + H2O(l) -> 2HNO3(l) ΔH° = -76.2 kJ (Equation 1)
H2O(l) -> H2(g) + 1/2O2(g) ΔH° = +286.0 kJ (Equation 2)
1/2N2(g) + 3/2O2(g) + 1/2H2(g) -> HNO3(l) ΔH° = -174.0 kJ (Equation 3)
Step 2: Multiply Equation 1 by 2 to match the stoichiometric coefficients of N2O5 in the target equation:
2N2O5(g) + 2H2O(l) -> 4HNO3(l) ΔH° = -152.4 kJ (Equation 4)
Step 3: Multiply Equation 2 by 2 and Equation 3 by 5 to match the stoichiometric coefficients of O2 in the target equation:
5H2O(l) -> 5H2(g) + 5/2O2(g) ΔH° = +572.0 kJ (Equation 5)
5/2N2(g) + 15/2O2(g) + 5/2H2(g) -> 5HNO3(l) ΔH° = -870.0 kJ (Equation 6)
Step 4: Reverse Equation 6 to match the stoichiometric coefficients of N2 in the target equation:
5HNO3(l) -> 5/2N2(g) + 15/2O2(g) + 5/2H2(g) ΔH° = +870.0 kJ (Equation 7)
Step 5: Add Equations 4, 5, and 7 to get the overall balanced equation:
2N2O5(g) + 5H2O(l) -> 4HNO3(l) + 5H2(g) + 5/2O2(g) + 5/2N2(g) ΔH° overall = -152.4 kJ + 572.0 kJ + 870.0 kJ
= +1289.6 kJ
Therefore, the delta H° for the reaction 2N2O5(g) -> 2N2(g) + 5O2(g) is +1289.6 kJ.