See image | AITS & Test Series — Chem-Mantra

Name: JEE Advanced Chemistry
Price: 7999 INR
Availability: InStock

Question

See image

Answer: 352

💡 Solution & Explanation

H = 70.4 kJ mol–1 = 70400 J mol–1 S = 2(400) – (250 + 350) = 200 J K–1 mol–1 H T S   (as G = H - TS = 0 at equm) = 70400 352K 200 

💬

Still have doubts about this question?

Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes