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.
|
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
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 1 images
