We appreciate your visit to The normal boiling point of water is 100 0 C and its molar enthalpy of vaporization is 40 67 kJ mol What is the change. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
ΔS= nΔHvap/T,
Where, ΔS = Change in entropy, n = moles of water = 39.3/18 = 2.188 moles, ΔHvap = 40.67kJ/mol = 40670 J/mol, T = Temperature (K) = 100+272.15 = 373.15 K
Therefore,
ΔS = (2.188*40670)/373.15 = 237.96 J/K
Where, ΔS = Change in entropy, n = moles of water = 39.3/18 = 2.188 moles, ΔHvap = 40.67kJ/mol = 40670 J/mol, T = Temperature (K) = 100+272.15 = 373.15 K
Therefore,
ΔS = (2.188*40670)/373.15 = 237.96 J/K
Thanks for taking the time to read The normal boiling point of water is 100 0 C and its molar enthalpy of vaporization is 40 67 kJ mol What is the change. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
