v4leriaaa

18.01.2021 •

Chemistry

n a closed system (N is constant) , a gas is taken from a pressure of 3.5atm to 1.75atm with the volume remaining constant (Isochoric) at 4.1L. In a second step, the gas expands at constant pressure (Isobaric) from a volume of 4.1L to 8.2L where the temperature reaches its original value. Show the process on a P vs. V What is the percentage change in Thermal Energy in the first step of the process

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Ответ:

jordanrose98

04.08.2022

Chemistry

$3\times 10^{35}$

Explanation:

The balanced chemical equation will be:

$2Ag(s)+Ni^{2+}(aq)\rightarrow 2Ag^{+}(aq)+Ni(s)$

Here Ag undergoes oxidation by loss of electrons, thus act as anode. Nickel undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

Where both E^0 are standard reduction potentials.

$E^0_{[Ag^{+}/Mg]}=+0.80V$

$E^0_{[Ni^{2+}/Ni]}=-0.25V$

$E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Ag^{+}/Ag]}$

E^0=-0.25-(+0.80V)=-1.05V

The standard emf of a cell is related to Gibbs free energy by following relation:

$\Delta G=-nFE^0$

$\Delta G$ = gibbs free energy

n= no of electrons gained or lost =?

F= faraday's constant

E^0 = standard emf

$\Delta G=-2\times 96500\times (-1.05)=202650J$

The Gibbs free energy is related to equilibrium constant by following relation:

$\Delta G=-2.303RTlog K$

R = gas constant = 8.314 J/Kmol

T = temperature in kelvin = 25^0C=25+273=298K

K = equilibrium constant

$\Delta G=-2.303RTlog K$

$+202650=-2.303\times 8.314\times 298\times logK$

$K=3\times 10^{35}$

Thus the value of the equilibrium constant at 25^0C is $3\times 10^{35}$

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