1
Question
The equilibrium constant for the reaction given below is 2.0×10−7 at 300 K. Calculate the standard entropy change if △H∘=28.40 kJmol−1 for the reaction:
PCl5(g)⇌PCl3(g)+Cl2(g)
The equilibrium constant for the reaction given below is 2.0×10−7 at 300 K. Calculate the standard entropy change if △H∘=28.40 kJmol−1 for the reaction:
PCl5(g)⇌PCl3(g)+Cl2(g)
Open in App
Solution
The correct option is A −33.6 Jmol−1K−1
K=2.0×10−7T=300K△H∘=28.4kJmol−1△G∘=−2.303 RT logk=−2.303×8.314×10−3×300×log(2×10−7)=−5.744(log2−7)=−5.744(0.3010−7)=38.48kJ
From △G∘=△H∘−T△S∘⇒T△H∘−△G∘△S∘=△H∘−△G∘T=28.4−38.48300=0.0336 kJ mol−1
or △S=−33.6 J mol−1K−1
K=2.0×10−7T=300K△H∘=28.4kJmol−1△G∘=−2.303 RT logk=−2.303×8.314×10−3×300×log(2×10−7)=−5.744(log2−7)=−5.744(0.3010−7)=38.48kJ
From △G∘=△H∘−T△S∘⇒T△H∘−△G∘△S∘=△H∘−△G∘T=28.4−38.48300=0.0336 kJ mol−1
or △S=−33.6 J mol−1K−1
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
