1.

Calculate the enthalpy change accompanying the transformation of C (graphite) to C(diamond). Given that the enthalpies of combustion of graphite and diamond are 393.5 and 395.4 kJ mol–1 respectively.

Answer»

Remember, enthalpy of combustion is always negative we are given

(i) C (graphite) + O2 (g) → CO2 (g) ; Δc H° = – 393.5 kJ mol–1

(ii) C (diamond) + O2 (g) → CO2 (g) ; Δc H°  = – 395.4 kJ mol–1

We aim at C(graphite) → C(diamond), Δtrans H°= ?

Subtracting eqn. (ii) from eqn. (i), we get

 C(graphite – C(diamond) → 0;

Δr H° = – 393.5 – (–395.4) = + 1.9 kJ 

or C(graphite) → (diamond) ; Δtrans H = + 1.9 kJ



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