+1 vote
in Chemistry by (29.7k points)
edited by

∆Uθof combustion of methane is – X kJ mol–1 . The value of ∆Hθ is

(i) = ∆Uθ

(ii) > ∆Uθ

(iii) < ∆Uθ

(iv) = 0

1 Answer

+1 vote
by (127k points)
selected by
Best answer

Since ∆Hθ = ∆Uθ + ∆ngRT and ∆Uθ = –X kJ mol–1

∆Hθ = (–X) + ∆ngRT. ⇒ ∆Hθ < ∆Uθ 

Therefore, alternative (iii) is correct. 

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.