Question 731
CBSE Class 11 Chemistry · Thermodynamics · Enthalpy changes and Hess law
For CCl₄(g)→C(g)+4Cl(g), use ΔvapH°=30.5, ΔfH°[CCl₄(l)]=−135.5, atomisation C=715 and Cl₂=242 kJ mol⁻¹ to find
(a)
atomisation enthalpy
(b)
mean C–Cl bond enthalpy.
Solution to Question 731
Complete workingBuild gaseous atoms from the elements, subtract formation of gaseous CCl₄, and divide the atomisation enthalpy by four equivalent C–Cl bonds.
\[\begin{aligned}\Delta_fH(CCl_4,g) &= -135.5+30.5\\&= -105.0\end{aligned}\]
\[\begin{aligned}\Delta_fH[C(g)+4Cl(g)] &= 715+2(242)\\&= 1199\end{aligned}\]
\[\begin{aligned}\Delta H &= 1199-(-105)\\&= 1304\ \mathrm{kJ\,mol^{-1}}\end{aligned}\]
\[\begin{aligned}D(C-Cl) &= \frac{1304}{4}\\&= 326\ \mathrm{kJ\,mol^{-1}}\end{aligned}\]
Final answers by part
(a)
Atomisation ≈1304 kJ mol⁻¹
(b)
mean C–Cl bond enthalpy ≈326 kJ mol⁻¹.