JEE MainChemistryToughPhysical Chemistry

The dissociation pressure of silver oxide at 445ºC is 207 atm.
Calculate $\Delta$ Gº for the formation of 1 mole Ag_2O(s) from metal
and oxygen at this temperature. (log 207 = 2.315). Report your answer in
Kcal and round of the value to nearest integer.

Show answer and solution

T = 445ºC = 445 + 273 = 718K
P = 207atm
n = 1 mole

{Ag}_{2}O \rightleftharpoons 2Ag(s) + \frac{1}{2}O_{2}(g)

K_{p} = \sqrt{P_{O_{2}}}

= (207)^1/2 = 14.39

\Delta G^{\circ} = \Delta G^{0}

+ 2.303 Rt log K_eq
But at eq.

\Delta

G = 0

\Delta

G^0 = - 2.303 RT log K_eq.
But for formation

2Ag(s) + \frac{1}{2}O_{2}(g) \rightleftharpoons {Ag}_{2}O(s)

1 Mole

\Delta G^{0}

= - 2.303 × 8.312 × 718 log

\frac{1}{(207)^{1/2}}

\Delta G^{0} = \frac{2.303 \times 8.312 \times 718}{2}

× log⁡207

\Delta G^{0}

= 6872.17 log 207

\Delta G^{0}

= 15915.75 J

\Delta G^{0}

= 3789.46 Cal

\Delta G^{0}

= 3.789 KCal

\Delta G^{0}

= 3.8 KCal

Source: Grass-G / CH-24 Chemistry Paper 31 Dec..docx

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