# Calculate the ent

Given:

ΔvapH° (CCl4) = 30.5 kJ mol–1

ΔfH° (CCl4) = -135.5 kJ mol–1

ΔaH° (C) = 715.0 kJ mol–1

ΔaH° (Cl2) =242 kJ mol–1

We can write,

CCl4(l) CCl4(g); ΔH = 30.5 kJ mol–1 (1)

C(s) + 2Cl2(g) CCl4(l); ΔH = -135.5 kJ mol–1 (2)

C(s) C(g); ΔH = 715.0 kJ mol–1 (3)

Cl2(g) 2Cl(g); ΔH =242 kJ mol–1 (4)

To get the required reaction, CCl4(g) C(g) + 4 Cl(g)

We follow the steps given below:

Step 1: Add t the both reaction (1) and (2), we get

C(s) + 2Cl2(g) CCl4(l); ΔH = -135.5 kJ mol–1

CCl4(l) CCl4(g); ΔH = 30.5 kJ mol–1

C(s) + 2Cl2(g) CCl4(g); ΔH = -105 kJ mol–1 (5)

Step 2: Add the both reaction (3) and (4), we get

C(s) C(g); ΔH = 715.0 kJ mol–1

2Cl2 (g) 4Cl (g); ΔH = 484 kJ mol–1 [ 2×reaction 4]

C(s) + 2Cl2(g) C(g) + 4Cl(g); ΔH = 1199 kJ mol–1 (6)

Step 3: Subtract the both reaction (5) and (6), we get

C(s) + 2Cl2(g) C(g) + 4Cl(g); ΔH = 1199 kJ mol–1

CCl4(g) C(s) + 2Cl2(g); ΔH = -105 kJ mol–1

CCl4(g) C(g) + 4 Cl(g); ΔH = 1304 kJ mol–1

(formation of required reaction)

Thus, the enthalpy change for the process is 1304 kJ mol–1

Calculation the bond enthalpy of C-Cl in CCl4(g)

Bond enthalpy of C-Cl bond in CCl4 is equal to one fourth of the energy of dissociation of CCl4

Bond enthalpy of C – Cl in CCl4(g) =

As ΔH =1304 kJ mol–1

= 326 kJ mol-1

Note: Bond enthalpy is defined as the amount of energy required to break one mole of bond of a particular type between the atoms in the gaseous state, i.e., to separate the atoms in the gaseous state under 1 am pressure and the specified temperature is called bond enthalpy.

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