One mole of graphite and diamond were separately burnt to form $CO_{2}$ gas.
The $Δ H^{\circ}$ values of the two reactions are given below:
C_(graphite) + O₂(g) $⟶$ CO₂(g) ; $Δ H^{\circ} = - 393.5 k J$
C_(diamond) + O₂(g) $⟶$ CO₂(g) ; $Δ H^{\circ} = - 395.4 k J$
Choose the correct statement.

Question

One mole of graphite and diamond were separately burnt to form $CO_{2}$ gas.

The $Δ H^{\circ}$ values of the two reactions are given below:

C_(graphite) + O₂(g) $⟶$ CO₂(g) ; $Δ H^{\circ} = - 393.5 k J$

C_(diamond) + O₂(g) $⟶$ CO₂(g) ; $Δ H^{\circ} = - 395.4 k J$

Choose the correct statement.

TestHub · Accepted Answer

The enthalpy of combustion for graphite is -393.5 kJ, and for diamond is -395.4 kJ. A more negative enthalpy of combustion indicates higher energy content and less stability. Since graphite's combustion releases less energy, it is more stable than diamond.

Therefore, graphite is more stable than diamond.

Chemistry - Thermochemistry Question with Solution | TestHub

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