NCERT Class XI Chemistry Thermodynamics Solutions
© examsnet.com
Question : 15
Total: 22
Calculate the enthalpy change for the process, C C l 4 ( g ) C ( g ) + 4 C l ( g ) C C l 4 ( g )
Δ v a p H ° ( C C l 4 ) k J m o l – 1 , Δ f H ° ( C C l 4 ) k J m o l – 1 , Δ a H ° ( C ) k J m o l – 1 Δ a H ° Δ a H ° ( C l 2 ) k J m o l – 1
Solution:
given : C C l 4 ( g ) C ( g ) + 4 C l ( g ) Δ v a p H ° ( C C l 4 ) k J m o l – 1
C ( s ) + 2 C l 2 ( g ) C C l 4 ( l ) Δ f H ° ( C C l 4 ) k J m o l – 1
C ( s ) C ( g ) Δ a H ° k J m o l − 1
C l 2 ( g ) 2 C l ( g ) Δ a H ° j m o l − 1
Required equation is :C C l 4 ( g ) C ( g ) + 4 C l ( g )
From Hess’s law,
eqn.(iii) + 2 × eqn.(iv) – eqn.(i) – eqn.(ii) gives required equation :
∴ ΔH = 715.0 + 2(242) – 30.5 – (–135.5) = 1304k J m o l – 1
Bond enthalpy of C – Cl inC C l 4
k J m o l − 1
Required equation is :
From Hess’s law,
eqn.(iii) + 2 × eqn.(iv) – eqn.(i) – eqn.(ii) gives required equation :
∴ ΔH = 715.0 + 2(242) – 30.5 – (–135.5) = 1304
Bond enthalpy of C – Cl in
© examsnet.com
Go to Question: