Q. 154.2( 33 Votes )

Calculate the ent

Answer :

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.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses