Question

14- Calculate AH-m for the following reaction: 4 NHs (g) 솨,e,s ? + 5 (g) → 4 NO (g) + 6 H2O (g) Given the following ΔΗ7 values: NHs (g)-45.9 kJ/mol NO (g) +91.3 kJ/mol H20 (g)-241.8 kJ/mol A) 1632 kJ B) 1632 kJ C) 1269 kJ D) 902 kJ E) Δ1p value for O2 is needed 15- Which electron configuration does not represent the Pauli exclusion A) B) s 2 C) 1s 2 2 D) 1s 2s E) 1s 2s 2

Answer 1

14. delta H_rxn= (product) - (reactant)

= [(4* 91.3) + (6 *-241.8)] - [ 4* -45.9] kJ/mol

= -902 kJ/mol (option D)

15. option E as the pair of electrons in 2s orbital have same spin and hence all the four same quantum number..which is not possible. So this configuration does not obey pauli exclusion principle.