Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
Bartleby Related Questions Icon

Related questions

Question
9:37:09
24.0 YA :49|39%
expert.chegg.com/exper
= Chegg
Time remaining: 01:50:25
Chemistry
What is the free energy change associated with the passage of a pair of electrons from
NADH to Oz if the passage of one proton from the matrix to the intermembrane space costs
+20.1 kJ mol-1 as shown in Sample Calculation 15.3? Use 3 significant figures.
AG=
kJ mol-1
SAMPLE CALCULATION 15,3
Problem
Calculate the free energy change for translocating a proton out of the mitochondrial matrix, where
pHmatriy = 7.8, pHtrol = 7.15, Ay = 170 mV, and T= 25°C.
Solution
Since pH = -log (H (Equation 2.4), the logarithmic term of Equation 15.7 can be rewritten.
Equation 15.7 then becomes
AG = 2.303 RT(pH - pH) + ZFAy
Substituting known values gives
AG = 2.303(8.3145 J K-- mol-)(298 K)(7.8 - 7.15)
+(1)(96,485 J- v-. mol(0.170 V)
= 3700 J- mol- + 16,400 J- mol-
= +20.1 kJ mol
Your answer
Typed answers are easier for students to read
than handwritten notes
Submit
Skip
expand button
Transcribed Image Text:9:37:09 24.0 YA :49|39% expert.chegg.com/exper = Chegg Time remaining: 01:50:25 Chemistry What is the free energy change associated with the passage of a pair of electrons from NADH to Oz if the passage of one proton from the matrix to the intermembrane space costs +20.1 kJ mol-1 as shown in Sample Calculation 15.3? Use 3 significant figures. AG= kJ mol-1 SAMPLE CALCULATION 15,3 Problem Calculate the free energy change for translocating a proton out of the mitochondrial matrix, where pHmatriy = 7.8, pHtrol = 7.15, Ay = 170 mV, and T= 25°C. Solution Since pH = -log (H (Equation 2.4), the logarithmic term of Equation 15.7 can be rewritten. Equation 15.7 then becomes AG = 2.303 RT(pH - pH) + ZFAy Substituting known values gives AG = 2.303(8.3145 J K-- mol-)(298 K)(7.8 - 7.15) +(1)(96,485 J- v-. mol(0.170 V) = 3700 J- mol- + 16,400 J- mol- = +20.1 kJ mol Your answer Typed answers are easier for students to read than handwritten notes Submit Skip
Expert Solution
Check Mark
Knowledge Booster
Background pattern image
Similar questions
Recommended textbooks for you
Text book image
Introduction to Chemical Engineering Thermodynami...
Chemical Engineering
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:McGraw-Hill Education
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemical Engineering
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Text book image
Elements of Chemical Reaction Engineering (5th Ed...
Chemical Engineering
ISBN:9780133887518
Author:H. Scott Fogler
Publisher:Prentice Hall
Text book image
Process Dynamics and Control, 4e
Chemical Engineering
ISBN:9781119285915
Author:Seborg
Publisher:WILEY
Text book image
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:9781285061238
Author:Lokensgard, Erik
Publisher:Delmar Cengage Learning
Text book image
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:9780072848236
Author:Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:McGraw-Hill Companies, The