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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

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

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:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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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
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
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