Question

1) Calculate the value of the Gibbs Free Energy at 298 K for the conversion of A to B if the enthalpy of the reaction is -5.75 kJ·mol-1 and the change in entropy is -26.2 J·K-1·mol-1: 2)For the conversion of reactant A to product B, the change in enthalpy is 3.67 kJ·mol-1 and the change in entropy is 84.8 J·K-1·mol-1. Above what temperature, in Kelvin, does the reaction become spontaneous?

          1) Calculate the value of the Gibbs Free Energy at 298 K
for the conversion of A to B if the enthalpy of the reaction is
-5.75 kJ·mol-1 and the change in entropy is -26.2
J·K-1·mol-1:
2)For the conversion of reactant A to product B, the change in
enthalpy is 3.67 kJ·mol-1 and the change in entropy
is 84.8 J·K-1·mol-1. Above
what temperature, in Kelvin, does the reaction become
spontaneous?
        
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Added by Antonio N.

Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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1) Calculate the value of the Gibbs Free Energy at 298 K for the conversion of A to B if the enthalpy of the reaction is -5.75 kJ·mol-1 and the change in entropy is -26.2 J·K-1·mol-1: 2)For the conversion of reactant A to product B, the change in enthalpy is 3.67 kJ·mol-1 and the change in entropy is 84.8 J·K-1·mol-1. Above what temperature, in Kelvin, does the reaction become spontaneous?
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Transcript

-
0:00 Hello students.
00:01 Here the question is in regarding to how to find that gives a free energy change of a reaction in which a is converted to b.
00:10 For this, the change in molar enthalpy of the reaction is given as here for this it is given in terms of kilojoules.
00:19 We need to convert it into joules and it is given as minus 5 .75 kilojoules molylers.
00:27 We have to convert into jols by multiplying the.
00:30 10 to the curve 3 jowl inverse and the change in molar and op is given as minus 26 .2 jalve kelvin inverse mole inverse and the temperature given at 2 98 kelvin.
00:49 Now we have to find out the gibbs free energy change.
00:53 For this the relation will be gibbs energy change is equal to delta h now minus t deltides.
01:01 Let us calculate it by putting the values which is equal to minus 5 .75 krog zones we have to move can able to convert into it.
01:12 Downs, mole inverse minus temperature 298 kelvin into it.
01:22 It will be minus 26 .2 joules kelvin inverse and mole inverse and from this we can be able to calculate now which we will get which is equal to minus 5 .75 into 10 to the power 3 jaw no inverse plus here can be cannabis so that by multiplying these two we will get plus to 780 746 jow no inverse from which we will get minus 5757.
02:08 Jullimbers plus 7807 .806 jull 0 .0 .0...
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