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DeltaG=DeltaH-TDeltaS and DeltaG=Delta...

`DeltaG=DeltaH-TDeltaS` and `DeltaG=DeltaH+T[(d(DeltaG))/(dT)]_(p),` then` ((dE_(cell))/(dT))` is

A

`(DeltaH)/(nF)`

B

`(DeltaG)/(nF)`

C

`(DeltaS)/(nF)`

D

`-(DeltaS)/(nF)`

Text Solution

AI Generated Solution

To solve the problem, we will derive the expression for \(\frac{dE_{cell}}{dT}\) using the relationships provided for \(\Delta G\). ### Step-by-Step Solution: 1. **Start with the given equations:** We have two expressions for \(\Delta G\): \[ \Delta G = \Delta H - T \Delta S \quad \text{(1)} ...
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DeltaG=DeltaH-TDeltaS and ltBRgt DeltaG=DeltaH+T[(d(DeltaG))/(dT)]_(p), then ((dE_(cell))/(dT)) is

Derive the equation, DeltaG = DeltaH - T DeltaS .

What is significance of T DeltaS in DeltaG = DeltaH - T DeltaS ?

The cell potential (E_(cell)) of a reaction is related as /_\G=-nF E_(cell) , where /_\G represents max. useful electrical work n=no. of moles of electrons exchanged during the section for reversible cell reaction d(/_\G)=(/_\_rV)dp-(/_\_rS),dT at constant pressure d(/_\G)=-(/_\_rS).dT :' At constant pressure /_\G=/_\H-T./_\S :. /_\G=H+T(d(/_\G))/((dT)_P) ((dE_(cell))/(dT))_P is known as temperture coefficient of the e.m.f of the cell. The temperature coefficient of the e.m.f of cell, ((dE)/(dT))_P si given by:

Gibbs-Helmoholtz equation relates the free energy change to the enthalpy and entropy changes of the process as (DeltaG)_(PT) = DeltaH - T DeltaS The magnitude of DeltaH does not change much with the change in temperature but the energy factor T DeltaS changes appreciably. Thus, spontaneity of a process depends very much on temperature. A reaction has value of DeltaH = 20 kcal at 200K , the reaction is spontaneous, below this temperature, it is not. the values DeltaG and DeltaS at 200K are, respectively

Gibbs-Helmoholtz equation relates the free energy change to the enthalpy and entropy changes of the process as (DeltaG)_(PT) = DeltaH - T DeltaS The magnitude of DeltaH does not change much with the change in temperature but the enrgy factor T DeltaS changes appreciably. Thus, spontaneity of a process depends very much on temperature. For the reaction at 298K, 2A +B rarr C DeltaH = 100 kcal and DeltaS = 0.020 kcal K^(-1) . If DeltaH and DeltaS are assumed to be constant over the temperature range, at what temperature will the reaction become spontaneous?

Gibbs-Helmoholtz equation relates the free energy change to the enthalpy and entropy changes of the process as (DeltaG)_(PT) = DeltaH - T DeltaS The magnitude of DeltaH does not change much with the change in temperature but the energy factor T DeltaS changes appreciably. Thus, spontaneity of a process depends very much on temperature. For the reaction at 25^(0)C, X_(2)O_(4)(l) rarr 2XO_(2) DeltaH = 2.0 kcal and DeltaS = 20 cal K^(-1) . the reaction would be

Free energy, G=H-TS, is a state function that includes whether a reaction is spontaneous or non-spontaneous. If you think of TS as the part of the system's energy that is disordered already, then (H-TS) is the part of the system's energy that is still ordered and therefore free to cause spontaneous change by becoming disordered. Also, DeltaG=DeltaH-TDeltaS To see what this equation for free energy change has to do with spontaneity let us return to relationship. DeltaS_("total")=DeltaS_("sys")+DeltaS_("surr") = DeltaS + DeltaS_("surr") (It is generally understood that symbols without subscript refer to the system not the surroundings.) DeltaS_("surr")=-(DeltaH)/T , where DeltaH is the heat gained by then system at constant pressure. DeltaS_("total") = DeltaS -(DeltaH)/T rArr TDeltaH_("total")=DeltaH-TDeltaS rArr -TDeltaS_("total") =DeltaH-TDeltaS i.e. DeltaG=-TDeltaS_("total") From second law of thermodynamics, a reaction is spontaneous if DeltaS_("total") is positive, non-spontanous if DeltaS_("total") is negative and at equilibrium if DeltaS_("total") is zero. Since, -TDeltaS=DeltaG and since DeltaG and DeltaS have opposite signs, we can restate the thermodynamic criterion for the spontaneity of a reaction carried out at constant temperature and pressure. If DeltaG lt 0 , the reaction is spontaneous. If DeltaG gt 0 , the reaction is non-spontanous. If DeltaG=0 , the reaction is at equilibrium. In the equation, DeltaG=DeltaH-TDeltaS , temperature is a weighting factor that determine the relative importance of enthalpy contribution to DeltaG . Read the above paragraph carefully and answer the following questions based on above comprehension: For the spontaneity of a reaction, which statement is true?

Free energy, G=H-TS, is a state function that includes whether a reaction is spontaneous or non-spontaneous. If you think of TS as the part of the system's energy that is disordered already, then (H-TS) is the part of the system's energy that is still ordered and therefore free to cause spontaneous change by becoming disordered. Also, DeltaG=DeltaH-TDeltaS To see what this equation for free energy change has to do with spontaneity let us return to relationship. DeltaS_("total")=DeltaS_("sys")+DeltaS_("surr") = DeltaS + DeltaS_("surr") (It is generally understood that symbols without subscript refer to the system not the surroundings.) DeltaS_("surr")=-(DeltaH)/T , where DeltaH is the heat gained by then system at constant pressure. DeltaS_("total") = DeltaS -(DeltaH)/T rArr TDeltaH_("total")=DeltaH-TDeltaS rArr -TDeltaS_("total") =DeltaH-TDeltaS i.e. DeltaG=-TDeltaS_("total") From second law of thermodynamics, a reaction is spontaneous if DeltaS_("total") is positive, non-spontanous if DeltaS_("total") is negative and at equilibrium if DeltaS_("total") is zero. Since, -TDeltaS=DeltaG and since DeltaG and DeltaS have opposite signs, we can restate the thermodynamic criterion for the spontaneity of a reaction carried out at constant temperature and pressure. If DeltaG lt 0 , the reaction is spontaneous. If DeltaG gt 0 , the reaction is non-spontanous. If DeltaG=0 , the reaction is at equilibrium. In the equation, DeltaG=DeltaH-TDeltaS , temperature is a weighting factor that determine the relative importance of enthalpy contribution to DeltaG . Read the above paragraph carefully and answer the following questions based on above comprehension: Which of the following is true for the reaction? H_(2)O(l) H_(2)O(g) at 100^(@) C and 1 atmosphere?

Free energy, G=H-TS, is a state function that includes whether a reaction is spontaneous or non-spontaneous. If you think of TS as the part of the system's energy that is disordered already, then (H-TS) is the part of the system's energy that is still ordered and therefore free to cause spontaneous change by becoming disordered. Also, DeltaG=DeltaH-TDeltaS To see what this equation for free energy change has to do with spontaneity let us return to relationship. DeltaS_("total")=DeltaS_("sys")+DeltaS_("surr") = DeltaS + DeltaS_("surr") (It is generally understood that symbols without subscript refer to the system not the surroundings.) DeltaS_("surr")=-(DeltaH)/T , where DeltaH is the heat gained by then system at constant pressure. DeltaS_("total") = DeltaS -(DeltaH)/T rArr TDeltaH_("total")=DeltaH-TDeltaS rArr -TDeltaS_("total") =DeltaH-TDeltaS i.e. DeltaG=-TDeltaS_("total") From second law of thermodynamics, a reaction is spontaneous if DeltaS_("total") is positive, non-spontanous if DeltaS_("total") is negative and at equilibrium if DeltaS_("total") is zero. Since, -TDeltaS=DeltaG and since DeltaG and DeltaS have opposite signs, we can restate the thermodynamic criterion for the spontaneity of a reaction carried out at constant temperature and pressure. If DeltaG lt 0 , the reaction is spontaneous. If DeltaG gt 0 , the reaction is non-spontanous. If DeltaG=0 , the reaction is at equilibrium. In the equation, DeltaG=DeltaH-TDeltaS , temperature is a weighting factor that determine the relative importance of enthalpy contribution to DeltaG . Read the above paragraph carefully and answer the following questions based on above comprehension: If an endothermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then

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