Calculate ΔH ° in kJ / mol for the reaction: NH3 (g) + Cl2 (g) → NH2Cl + HCl (g) based on the energy of tenons that break and form in the reaction. The bond enthalpies of the following compounds are given: N-H (389), Cl-Cl (243), N-Cl (201), and H-Cl (431) in kJ / mol
B= bond energy. See the number of different bonds in the molecules.
Check the calculations.
Calculate ΔH ° in kJ / mol for the reaction: NH3 (g) + Cl2 (g) → NH2Cl + HCl (g) based on the energy of tenons that brea...
show work 9) Given the bond enthalpies Cl-Cl (243), C-CI (339), H-CI (431),C-H (414)all in kJ/mol, estimate AE for the following reaction: CH4(g)+ Cl2(g)-CH3Cl(g)+ HCl(g) A)-1427 k) B) -113 kJ C) -30 kJ D)-356 kJ E) 301 kJ
Use the ΔH°f values provided to determine ΔH°rxn for the following reaction CH4(g) + 3 Cl2(g) → CHCl3(l) + 3 HCl(g) ΔH°rxn = ? ΔH°f (kJ/mol): -75 -134 -92 A. +662 kJ B. +117 kJ C. -151 kJ D. -335 kJ The equation that corresponds to the enthalpy of formation for NH3(g) is A. N(g) + 3H(g) → NH3(g) B. N(g) + 3/2 H2(g) → NH3(g) C. 1/2 N2(g) + 3H(g) → NH3(g) D. 1/2 N2(g) + 3/2 H2(g) →...
Consider the following information. The lattice energy of LiCl is ΔH lattice = −834 kJ/mol. The enthalpy of sublimation of Li is ΔH sub = 159.3 kJ/mol. The first ionization energy of Li is IE 1 = 520 kJ/mol. The electron affinity of Cl is ΔH EA = -349 kJ/mol. The bond energy of Cl2 is BE = 243 kJ/mol. Determine the enthalpy of formation, ΔHf, for LiCl(s).
Use the bond energies provided to estimate ΔH°rxn for the reaction below.PCl3(g) + Cl2(g) → PCl5(l)ΔH°rxn = ?BondBond Energy (kJ/mol)Cl-Cl243P-Cl331
Consider the following information. The lattice energy of NaCl is ΔH lattice=−788 kJ/mol The enthalpy of sublimation of Na is ΔHsub=107.5 kJ/mol The first ionization energy of Na is IE1=496 kJ/mol. The electron affinity of Cl is ΔHEA=−349 kJ/mol. The bond energy of Cl2 is BE=243 kJ/mol. Determine the enthalpy of formation, ΔHf, for NaCl(s). ΔHf= kJ/mol
Calculate the ΔH0 for the following reaction: NH4Cl(s) → NH3(g) + HCl(g) Given the following standard enthalpies of formation ΔHf0 (298 K, 1 atm) NH3(g) -46.2 kJ mol-1 ; HCl(g) -92.3 kJ mol-1 ; NH4Cl(s) -315.0 kJ mol-1
The standard enthalpy of a reaction, AH, is the energy or enthalpy change associated with transforming reactants into products, where "standard", denoted °, implies reaction conditions of 25°C, 1 atm and/or 1 M. Standard enthalpies of a reaction can be estimated by assuming that all reactant bonds are broken before forming all product bonds: AHTD =( enthalpies of bonds broken) - (2 enthalpies of bonds formed) Using the information below, calculate AH°r (kJ) for the reaction of H2(g) + Cl2(g)...
calculate the lattice energy of NaCl based on the given information : ΔH°f[NaCl(s)] = -411 kJ/mol ΔH°f [Clg] = 121.5 kJ/mol ΔH°sublimation [Na] = 109 kJ/mol IE1 (Na) = 496 kJ/mol EA1 (Cl) = -349 kJ/mol
Part A. Given the bond dissociation energies (in kJ/mol) for the following diatomic molecules Cl2 (243), F2 (158), H2 (436), O2 (498), N2 (945) choose the one(s) that could be broken by using blue light (λ=465 nm). Part B. Given the bond energies (in kJ/mol) of the following bonds: F–F (155), F–Cl (193), and Cl–Cl (243), estimate the molar enthalpy of formation of ClF(g), that is find ∆H for the following reaction ½Cl2(g) + ½F2(g) → ClF(g)
1. a. Use the bond energy data provided to calculate AH.. for the following reaction. H-H + Cl-Cl → 2 H-CI Bond Bond Energy H-H 436 kJ/mol Cl-CI 243 kJ/mol H-CI 431 kJ/mol b. Based on your answer to part a, is the reaction an exothermic reaction or an endothermic reaction? Provide a brief explanation to support you choice of answer. An endothermic reaction occurs in a cylinder with initial volume of 1 L. 5 kJ of heat is transferred...