5. Given the following data: 2 H2(g) + O2(g) → 2 H20 (1) AH° = -571.6 kJ N,Os (g) + H20 (1) ► 2 HNO (1) AH = -76.6 kJ N2(g) + 3 O2 (g) + H2(g) → 2 HNO, (1) AH = -348.2 kJ a. Calculate the AHⓇ for the reaction: 2 N2 (g) + 5 O2(g) → 2 N2Os (g)
Consider the following data. 2 H2(g) + O2(g) 2 H2O(l) ΔH = -571.7 kJ N2O5(g) + H2O(l) 2 HNO3(l) ΔH = -92.0 kJ N2(g) + 3 O2(g) + H2(g) 2 HNO3O(l) ΔH = -348.2 kJ Use Hess's law to calculate ΔH for the reaction below. 2 N2O5(g) 2 N2(g) + 5 O2(g) ΔH = _____kJ
Given the following equations and Ho values given below, determine the heat of reaction at 298 K for the reaction: 2 N2(g) + 5 O2(g) 2 N2O5(g) 2 H2(g) + O2(g) 2 H2O(l) Ho/kJ = -571.6 N2O5(g) + H2O(l) 2 HNO3(l) Ho/kJ = -73.7 N2(g) + 3 O2(g) + H2(g) 2 HNO3(l) Ho/kJ = -348.2
You are given the following data. AH = = -286.0 kJ H2(g) + 1/2O2(g) → H20(1) N2O5(9) + H20(1) → 2 HNO3(aq) 1/2 N2(9) + 3/2 02(9) + 1/2 H2(9) AH = -77.0 kJ HNO3(1) AH = -174.0 kJ Calculate the AH for the reaction given below. 2 N2(g) + 5 O2(g) → 2 N205(9) k] Need Help? Read It Supporting Materials Periodic Table E Constants & Factors Supplemental Data
Calculate the standard enthalpy of formation of liquid water (H2O) using the following thermochemical information: 2 HNO3(l) N2O5(g) + H2O(l) H = +92.0 kJ 2 N2(g) + 5 O2(g) 2 N2O5(g) H = +59.3 kJ N2(g) + 3 O2(g) + H2(g) 2 HNO3(l) H = -348.2 kJ
From the following equations, calculate AH for the reaction 2N2(g) + + 5O2(g) → 2N2O5(g) AH = ? H2(g) + O2(9) H206) AH = -285.8 kJ N2O5(g) + H200 2HNO30 AH = -76.6 kJ N2(g) + 302(g) + H219) - 2HNO30) AH = -348.2 kJ
Calculate AH for the reaction N2H4(0) + O2(g) → N2(g) + 2 H2O(1) given the following data: Equation AH (kJ) 2 NH3(g) + 3 N2O(g) → 4 N2(g) + 3 H2O(l) -1010 N2O(g) + 3 H2(g) → N2H4(1) +H2O(1) -317 2 NH3(g) + 1202(g) → N2H4(1) + H20(1) -143 H2(g) + 1202(g) → H2O(H -286 AH=
2. Given the following data: H2O(l) → H2(g) + 1/2O2(g) ΔH° = 285.8 kJ 2HNO3(l) → N2O5(g) + H2O(l) ΔH° = 76.6 kJ 2N2(g) + 5O2(g) → 2N2O5(g) ΔH° = 28.4 kJ Calculate ΔH° for the reaction: 1/2N2(g) + 3/2O2(g) + 1/2H2(g) → HNO3(l) Note that you should be able to answer this one without needing to use any additional information from the thermo table. I've attempted this question multiple times. I am able to get to the simplified eqaution...
Using the reactions below, determine the ∆Hº for 2 N2(g) + 5 O2(g) --> 2 N2O5(g) N2(g) + 3 O2(g) + H2(g) --> 2 HNO3(aq) ∆Hº = -415 kJ/mol N2O5(g) + H2O(l) --> 2 HNO3(aq) ∆Hº = -140 kJ/mol 2 H2(g) + O2(g) --> 2 H2O(l) ∆Hº = -572 kJ/mol
Given that H2(g) + F,() 2 HF(g) AH = -546.6 kJ 2 H,() + O2(g) → 2H,O(1) AH x = -571.6 kJ calculate the value of AHixn for 2F2(g) + 2 H2O(1) 4 HF(g) + O2(8) AH = KJ