Q.15. Calculate the enthalpy change for the process

CCl4gCg+4Clg and calculate bond enthalpy of C-Cl in CCl4g.

vapHθCCl4=30.5kJmol-1.
fHθCCl4=135.5kJmol-1.
aHθC=715.0kJmol-1,whereaHθisenthalpyofatomisation
aHθCl2=242kJmol-1

NEETprep Answer:
The chemical equations implying to the given values of enthalpies are:
iCCl4lCCl4gvapHθ=30.5kJmol-1iiCsCgaθH=715.0kJmol-1iiiCl2g2ClgaθH=242kJmol-1ivCg+4ClgCCl4grH=-135.5kJmol-1
Enthalpy change for the given process CCl4gCg+4Clg can be calculated using the following algebraic calculations as:
Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)
H=aHθC+2aHθCl2-vapHθ-fH=715.0kJmol-1+2242kJmol-1-30.5kJmol-1--135.5kJmol-1H=1304kJmol-1
Bond enthalpy of C-Cl bond in CCl4g
13044kJmol-1=326kJmol-1