Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
80 views
in Chemistry by (70.3k points)
closed by
For oxidation of iron.
`4Fe(s)+3O_(2)(g)rarr2Fe_(2)O_(3)(s)`
entropy change is`- 549.4 JK^(-1)mol^(-1)` at 298 K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous?
`(Delta_(r)H^(Ө)` for this reaction is `-1648xx10^(3)J"mol"^(-1)`)

1 Answer

0 votes
by (59.2k points)
selected by
 
Best answer
One decides the spontaneity of a reaction by considering
`DeltaS_("total")(DeltaS_("sys")+DeltaS_(surr")).` For calculating `DeltaS_("surr")` have to consider the heat absorbed by the surroundings which is equal to `-Delta_(r)H^(Ө)` . At temperature T, entropy change of the surroundings is
`DeltaS_("surr")=-(Delta_(r)H^(Ө))/(T)` (at constant pressure)
`=-((-1648xx19^(3)J "mol"^(1)))/(298K)`
`5530 JK^(-1)"mol"^(-1)`
Thus, total entropy change for this reaction
`Delta_(r)S_("total")=5530JK^(-1)"mol"^(-1)+`
`" " (-549.4JK^(-1)"mol"^(-1))`
`=4980.6 JK^(-1)"mol"^(-1)`
This shows that the above reaction is spontaneous.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...