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).

ΔvapH?(CCl4) = 30.5 kJ mol–1.

ΔfH? (CCl4) = –135.5 kJ mol–1.

ΔaH? (C) = 715.0 kJ mol–1, where ΔaHis enthalpy of atomisation

ΔaH? (Cl2) = 242 kJ mol–1  

7 Views|Posted 8 months ago
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8 months ago

15. According to the question:

i.  CCl4  (l) → CCl4  (g),                        ? vapH? = 30.5 kJ mol–1

iii. C (s) + 2Cl2 (g)  CCl4  (l),              ? fH? = –135.5 kJ mol–1

iii. C (s) → C (g),                                  ? aH? = 715.0 kJ mol–1

iv Cl2  (g) → C (g) + 4 Cl (g)                 ? aH

...Read more

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Chemistry Ncert Solutions Class 11th 2023

Chemistry Ncert Solutions Class 11th 2023

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