Question

At 25 C:

PbO(s) ---> Pb(s) + ½ O2(g)

dH (kJ/mol) -217.32 0 0

dS (J/mol K) 68.7 64.81 205.138

dG (kj/mol) -187.89 0 0

Calculate dH for the previous reaction.

A. -98.68 kJ/mol

B. 98.68 kJ/mol

C. -217.32 kJ/mol

D. 217.32 kJ/mol

Answer 1

Given chemical transformation (Decomposition)

       PbO(s) ---> Pb(s) + ½ O2(g)

And given data,

ΔH (PbO) = -217.32 kJ, ΔH (Pb) = 0 kJ and ΔH (O₂) = 0 kJ

Let us calculate ΔH for given decomposition is,

ΔH = ∑ ΔH(products) - ∑ ΔH(reactant)

ΔH = [ΔH (Pb) + ½ ΔH (O₂)] –[ ΔH (PbO)]

ΔH = [(0) + ½ (0)] – [-217.32]

ΔH = 0 – (-217.32)

ΔH = +217.32 kJ.

Hence the enthalpy change for given decomposition (i.e. reaction in forward direction) is +217.32 kJ.

For reverse reaction i.e. formation of PbO from Pb and O2 the enthalpy change (ΔH) will be equal in magnitude but with reverse in sign I.e. -217.32 kJ.

Not: Standard Gibbs energy and standard entropy changes are not required. With those values also we will reach to the same answer. This extra data would be of use if standar enthalpy changes ΔH values if not provided or known.

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