Question

Use standard enthalpies of formation in Appendix L to calculate enthalphy changes for the following

(A) 1.0 g of white phosphorus burns, forming P4 010(s)
( b) 0.20 mol of NO (g) decompose to N2(g) and O2 (g)
(C) 2.40 g of NaCl s) is formored from Na(s) and excess Cl2(g)

Answer 1

he reaction enthalpy is the difference between the amount of enthalpies for the formation of reaction products and the amount of enthalpy formation of reagents, taking into account stoichiometric amounts of substances:

ΔH = Σ (nΔH)prod. – Σ (nΔH)react.

The enthalpies of the formation of simple substances are zero.

(A) P4 + 5O2 → P4O10(s) + 2984 kJ

ΔH°(P4O10) = - 2984 kJ/mol

n (P4) = 1g/31·4g/mol = 0.008 mol

1 mol P4 - (-2984) kJ

0.008 mol - ΔH

ΔH = 0.008·(-2984) = -23.87 kJ

( B) 2NO(g) → N2(g) + O2(g)

ΔH°(NO) = 90.3 kJ/mol

ΔH°reaction = - 90.3 kJ/mol

1 mol NO - (- 90.3) kJ

0.2 mol - ΔH

ΔH = 0.2·(-90.3) = -18.06 kJ

(C) 2Na(s) + Cl2(g) → 2NaCl(s)

ΔH°(NaCl) = -411.1 kJ/mol

n (NaCl) = 2.4 g/58.5g/mol = 0.041mol

1 mol NaCl - (-411.1) kJ

0.041 mol - ΔH

ΔH = 0.041·(-411.1) = 16.855 kJ