{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Exam2BKey

Exam2BKey - 1 Use the following thermodynamic data in this...

This preview shows pages 1–2. Sign up to view the full content.

1) Use the following thermodynamic data in this question. Please note the units for each answer! 2 C(g) + O 2 (g) ¾ 2 CO(g) H o f : 716.68 0 - 110.52 kJ/mol S o : 158.0 205.0 197.6 J/mol-K a) (15) What is the value of H o for this reaction to form two moles of CO(g)? H o = Σ∆ H o f (prod) - Σ H o f (reac) = [ 2 (-110.52) ] - [ 2 (716.68) + 0 ] = -1654 kJ b) (15) What is the value of S o for this reaction to form two moles of CO(g)? S o = Σ S o (prod) - Σ S o (reac) = [ 2 (197.6) ] - [ 2 (158.0) + 1 (205.0) ] = -125.8 J / K c) (15) What is the value of G o for this reaction to form two moles of CO(g) at 1775 K? G o = H o - T S o = (-1654 kJ) - (1250 K) (-0.1258 kJ/K) = -1431 kJ d) (15) What is the value of K at 1775 K? K = exp [- G o /RT] = exp [-(-1431)/(0.00831)(1775) = 1.36 x 10 42 e) (10) Based on your calculations, is this reaction spontaneous at 1775 K (yes or no)? yes f) (10) Based on your calculations, is this reaction extensive at 1775 K (yes or no)? yes g) (15) Is there a temperature at which the spontaneity reverses? If so, calculate that temperature. If not, write “No Temperature”.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern