wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

CCl4(g)C(g)+4C(g)
Calculate the bond enthalpy of CCl in CCl4.
ΔHvapCCl4=30.5KJmol1
ΔHfCCl4=135.5KJmol1
ΔHaC=715.0KJmol1
ΔHaCl2=242KJmol1

Open in App
Solution

The chemical equations implying to the given values of enthalpies are:
(i) CCl4(l)CCl4(g) ΔHvap:30.5kJ/mol
(ii) C(s)C(g) ΔHa=715.0kJ/mol
(iii) Cl2(g)2Cl(g) ΔHa=242kJ/mol
(iv) C(g)+4Cl(g)CCl4(g) ΔHf=135.5kJ/mol
Enthalpy change for the given process
CCl4(g)C(g)+4Cl(g)
Equation (ii) +2× Equation (iii) Equation (i) Equation (iv)
=H=1304kJ/mol
Bond enthalpy of CCl bond in CCl4(g)=13044
=326kJ/mol

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Bond Parameters
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon