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Determine $\Delta G$ and $\Delta G^{\circ}$ for each of the reactions in the previous problem.

a. $\Delta G(\text {standard})=10.613 \frac{k J}{m d}$$\Delta G=-12.542 \frac{k J}{m o l}$b. $\Delta G\left(\operatorname{atand}(x+d)=-813326 \frac{k l}{m a d}\right.$$\Delta G=10.420 \frac{k \cdot l}{m o l}$c. $\Delta G(s t a n d a r d)=1591.515 \frac{k . J}{m o l}$$\Delta G=-29.523 \frac{k J}{m d}$

Chemistry 102

Chapter 17

Electrochemistry

Carleton College

University of Central Florida

University of Kentucky

Brown University

Lectures

03:07

A liquid is a nearly incom…

04:38

A liquid is a state of mat…

01:37

Calculate $\Delta G^{\circ…

01:13

01:02

01:43

03:26

Using the following values…

10:30

Find $\Delta G^{\circ}$ fo…

09:10

Using values of $\Delta G_…

13:19

Predict the sign of $\Delt…

04:43

06:04

Calculate $\Delta H^{\circ…

05:47

Use standard free energies…

01:11

07:38

01:33

05:55

01:27

07:15

Calculate Grxn for the r…

04:46

Determine $\Delta G^{\circ…

02:01

06:41

normally to conclude the value off Delta Gino Story to calculate the value of Delta Gino, we applies this equation. The original is equal number off electron transfer times about the value off in on off the cell times by for the canister, which is the Faraday Constant is constant. So in a transfer times boy e not off this times, boy 96485 by applies is in the first bar, we found that the Delta Gino is equal negative. The number off transfer electoral in the question 57 is six and the, you know off this is 1.26 and the for today is 96485 while in case be delta Jeannot is equal to the number off electron transfer, which is toe and a valuable innocent is negative or 0.54 times by for a day. Conversant, which is 9648 fine. And the result will be Jewell bear more. This is a unit off the result. Thank you

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