LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
in Chemistry by (66.9k points)
closed by
Find the enthalpy of formation of hydrogen flouride on the basis of following data:
Bond energy of `H-H` bond `=434 kJ mol^(-1)`
Bond energy of `F-F` bond `=158 kJ mol^(-1)`
Bond enegry of `H -F` bond `=565 kJ mol^(-1)`

1 Answer

0 votes
by (70.0k points)
selected by
Best answer
`(1)/(2) H_(2)(g) +(1)/(2) F_(2)(g) rarr HF(g)`
`Delta_(f)H = BE` of reactant -`BE` of product
`= ((1)/(2)BE of H_(2) +(1)/(2)BE of H_(2)) - (BE of HF)`
`=(1)/(2) xx 434 +(1)/(2) xx 158 - 565 = 217 +79 - 565`
`=- 269 kJ`

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.