Here is the reaction of carbon monoxide and hydrogen to form gaseous methanol at 298 K. (a) Calculate DSo for this reaction at 298 K. DSo= (include units) This reaction is spontaneous at 298 K under standard conditions. Substance DHfo kJ/mol So J/(K mol) CH3OH (g) -201 239.9 H2 (g) 130.7 CO (g) -110.5 197.7 CO (g) + 2 H2 (g)  CH3OH (g) (b) Based on the particular reactants and products present, explain the sign of your answer for DSo. Give two separate reasons. (c) Find the value of DHo for this reaction under standard conditions at 298 K. DHo= (give units) (d) If you did the calculation in (a) correctly, you should have found out that DSo is negative. How can this reaction be spontaneous if DSo is negative? Your explanation must state and use the Second Law of Thermodynamics.

Here is the reaction of carbon monoxide and hydrogen to form gaseous methanol at 298 K. (a) Calculate DSo for this reaction at 298 K. DSo= (include units) This reaction is spontaneous at 298 K under standard conditions. Substance DHfo kJ/mol So J/(K mol) CH3OH (g) -201 239.9 H2 (g) 130.7 CO (g) -110.5 197.7 CO (g) + 2 H2 (g)  CH3OH (g) (b) Based on the particular reactants and products present, explain the sign of your answer for DSo. Give two separate reasons. (c) Find the value of DHo for this reaction under standard conditions at 298 K. DHo= (give units) (d) If you did the calculation in (a) correctly, you should have found out that DSo is negative. How can this reaction be spontaneous if DSo is negative? Your explanation must state and use the Second Law of Thermodynamics.

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter10: Entropy And The Second Law Of Thermodynamics

Section: Chapter Questions

Problem 10.33PAE: Limestone is predominantly CaCO3, which can undergo the reaction CaCO3(s)CaO(s)+CO2(g). We know from...

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Here is the reaction of carbon monoxide and hydrogen to form gaseous methanol at 298 K.

(a) Calculate DS^o for this reaction at 298 K.

DS^o=

(include units)

This reaction is spontaneous at 298 K under standard conditions.

Substance	DH_f^o kJ/mol	S^o J/(K mol)
CH₃OH (g)	-201	239.9
H₂ (g)		130.7
CO (g)	-110.5	197.7

CO (g) + 2 H₂ (g)  CH₃OH (g)

(b) Based on the particular reactants and products present, explain the sign of your answer for DS^o. Give two separate reasons.

DH^o=

(give units)

(d) If you did the calculation in (a) correctly, you should have found out that DS^o is negative. How can this reaction be spontaneous if DS^ois negative? Your explanation must state and use the Second Law of Thermodynamics.

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