Calculate the enthalpy change for the reaction
NO(g) + O(g) → NO2(g)
from the following data
NO(g) + O3(9) → NO2(g) + O2(g) ΔH=-198.9 kJ/mol
O3(g) → 1.5O26(g) ΔH=-142.3 kJ/mol
O2(g) → 2O(g) ΔH = 495.0 kJ/mol
A. 153.8 kJ B. 190.9 kJ C.-551.6 kJ D.-304.1 kJ E. 438.4 kJ
Calculate the enthalpy change for the reaction NO(g) + O(g) → NO2(g)
Question 62 Calculate the enthalpy change for the following reaction: NO(g) + O(g) - NO2(g) from the following data: NO(g) + O3(g) – NO2(g) + O2(g) O3(g) – 1.502(g) O2(g) - 20(g) AH = -198.9 kJ AH = -142.3 kJ AH = 495.0 kJ 0-304.1kJ 438.4 kJ 190.9 kJ 153.8 kJ O -551.6 kJ
5) Calculate the enthalpy change for the reaction NO(g)+ Og) NO2g) from the following data: NOg) +03(g)-NO2(g) + O2(g) ΔΗ--I 98.9 kJ O3g) 1.502(g) 02(g) 20(g) AH-142.3 kJ AH = 495.0 kJ A) 153.8 kJ B) 438.4 kJ C)-551.6 kJ D) 190.9 kJ E)-304.1 kJ
Calculate ΔH for the reaction: NO (g) + O (g) → NO2 (g) given the following information: NO(g) + O3(g) → NO2(g) + O2(g) ΔH = -198.9 kJ O3(g) → 3/2 O2(g) ΔH = -142.3 kJ O2(g) → 2 O(g) ΔH = 495.0 kJ I understand the basic concept for the problem, the 1/3 O2 is throwing me off. Thanks!
physical chemistry
2. The bond dissociation energy of oxygen gas is 495 kJ/mole, while the enthalpy of formation of ozone gas (O3(g))is 142.3 kJ/mole. For the reaction: NO +O3(g) NO2(g) + O2(g) it is found that AH = -198.9 kJ/mole, Determine the enthalpy change for the reaction: NO(g) + O() NO2(8)
I'm really focused on getting an explanation with this so I can do the rest of them, so if you could explain and answer that would be great. Given the following thermochemical equations, NO(g) + O3(g) → NO2(g) + O2(g) where Δ H = −198.9 kJ/mol O3(g) → 3 2 O2(g) where Δ H = −142.3 kJ/mol O2(g) → 2O(g) where Δ H = +495 kJ/mol Determine the enthalpy change for the reaction: 2NO2(g) → 2NO(g) + O2(g)
Calculate the enthalpy of the reaction: 2 NO(g) + O2(g) --> 2 NO2(g) given the following reactions and enthalpies of formation: 1/2 N2(g) + O2(g) → NO2(g), ΔH = 33.2 kJ 1/2 N2(g) + 1/2 O2(g) → NO(g), ΔH = 90.2 kJ
Consider the following chemical reaction. NH3(g) + 2 O2(g) → HNO3(aq) + H2O(l) Calculate the change in enthalpy (ΔH) for this reaction, using Hess' law and the enthalpy changes for the reactions given below. (1a) 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l); ΔH = −1166.0 kJ/mol (2a) 2 NO(g) + O2(g) → 2 NO2(g); ΔH = −116.2 kJ/mol (3a) 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g); ΔH = −137.3 kJ/mol
Part A - Calculating an Enthalpy of Reaction from Enthalpies of Formation Calculate the enthalpy change for the reaction: 2 H2O2(l) → 2 H2O(l) + O2(g) using enthalpies of formation: ΔH∘f[H2O2]ΔH∘f[H2O]==−187.8 kJ/mol−285.8 kJ/mol Calculate the enthalpy change for the reaction: using enthalpies of formation: Multiple choice answers below: -98.0 kJ -196.0 kJ +98.0 kJ +196.0 kJ
Calculate the standard enthalpy of formation of NO(g) (in kJ/mol) from the following data. N2(g) + 2 O2(g) → 2 NO2(g) ΔH⁰298 = 66.4 kJ 2 NO(g) + O2(g) → 2 NO2(g) ΔH⁰298 = −116.2 kJ
Given the following data:
2O3(g)
3O2(g)
-427 kJ
O2(g)
2O(g)
495 kJ
NO(g)
+
O3(g)
NO2(g)
+
O2
-199kJ
Calculate
(in kJ) for the reaction:
NO(g)
+
O(g)
NO2(g)