Question: (a) For CO2 a geometry-optimized HF/6-31G* calculation gives Ue = -187.634176 hartrees. UHF/6-31G* ground-state energies are -37.680860 and -74.783931 hartrees for C and O, respectively.
(b) Unscaled HF/6-31G* vibrational wavenumbers for CO2 are 745.8, 745.8, 1518.5, and 2585.0 cm-1. Scale these by 0.89, find the HF/6-31G* prediction for D0, and compare with the experimental value D0 = 16.56 eV. Also calculate the predicted atomization energy in kcal/mol.
(c) Thermodynamic tables give ∆Hof,0 as 246.79 kJ/mol for O(g) and 711.2 kJ/mol for C(g) and give the molar enthalpy difference Hm,298 - Hm,0 as 1.05 kJ/mol for graphite. Find the HF/6-31G* predictions for ∆Hof,0 and ∆Hof,298 of CO2(g). The experimental values are -393.2 kJ/mol and -393.5 kJ/mol.
Step by Step Solution
3.40 Rating (166 Votes )
There are 3 Steps involved in it
a D e hartrees 2 74783931 37680860 187634176 0385454 D e 0385454 hartree272114 eVhartree 10489 eV b ... View full answer
Get step-by-step solutions from verified subject matter experts
Document Format (1 attachment)
959-P-M-A-M (871).docx
120 KBs Word File
