15. Calculate the enthalpy change for the process

CCl₄(g) → C(g) + 4 Cl(g)

and calculate bond enthalpy of C – Cl in CCl₄(g).

Δ_vapH^?(CCl₄) = 30.5 kJ mol–1.

Δ_fH^? (CCl₄) = –135.5 kJ mol–1.

Δ_aH^? (C) = 715.0 kJ mol–1, where Δ_aH^? is enthalpy of atomisation

Δ_aH^? (Cl₂) = 242 kJ mol–1  

<p><strong>15.&nbsp;</strong>According to the question:</p><p>i.&nbsp; CCl₄&nbsp; (l) → CCl₄&nbsp; (g), &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;? _vapH? = 30.5 kJ mol^–1</p><p>iii. C (s) + 2Cl₂ (g)&nbsp;<strong>→</strong>&nbsp;CCl₄&nbsp; (l), &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;? _fH? = –135.5 kJ mol^–1</p><p>iii. C (s)&nbsp;<strong>→&nbsp;</strong>C (g), &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;? _aH? = 715.0 kJ mol^–1</p><p>iv Cl₂&nbsp; (g) → C (g) + 4 Cl (g)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;? _aH? = 242 kJ mol^–1</p><p>The equation we want is:</p><p>CCl_4&nbsp; (g) → C (g) + 4Cl (g), &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;? H =? </p><p>Equation (iii) + {2 x Equation (iv)} – Equation (ii), gives the required equation with</p><p>? H = 715.0 + 2 x (242) – 30.5 – (-135.5) kJ mol^-1</p><p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= 1340 kJ mol^-1</p><p>Bond enthalpy of C-Cl in CCl₄&nbsp;= 1304 / 4 = 326 kJ mol^-1</p>

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