The dissociation pressure of silver oxide at `445^(@)C "is" 207` atm. Calculate `DeltaG^(@)` for the formation of `1` mole `Ag_(2)O(S)` from metal and oxygen at this temperature. (`"log" 207=2.315`)

The dissociation pressure of silver oxide at `445^(@)C "is" 207` atm.

1.	The dissociation pressure of silver oxide at `445^(@)C "is" 207` atm. Calculate `DeltaG^(@)` for the formation of `1` mole `Ag_(2)O(S)` from metal and oxygen at this temperature. (`"log" 207=2.315`)
Answer» Correct Answer - `3.8Kcal` `T=445^(@)C=445+273=718K` `P=207`atm `n=1` mole `Ag_(2)OhArr2Ag(s)+(1)/(2)O_(2)(g)` `K_(p)=sqrtPo_(2)=(207)^(1//2)=14.39` `DeltaG^(@)=DeltaG^(@)+2.303RT"log"K_(eq.)` But at eq. `DeltaG=0` `DeltaG^(@)=-2.303RT"log" K_(eq.)` But for formation `2Ag(S)+(1)/(2)O_(2)(g)hArrAg_(2)O(S)` `1` mole `K_(P)=(1)/((Po_(2))^(1//2))=(1)/((207)^(1//2)` `DeltaG^(@)=-(2.303xx8.312xx718)/(2)=xx"log"207` `DeltaG^(@)=6872.17log 207` `DeltaG^(@)=15915.75J` `DeltaG^(@)=3789.46 Cal` `DeltaG^(@)=3.789 KCal` `DeltaG^(@)=3.8KCal`

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The dissociation pressure of silver oxide at `445^(@)C "is" 207` atm. Calculate `DeltaG^(@)` for the formation of `1` mole `Ag_(2)O(S)` from metal and oxygen at this temperature. (`"log" 207=2.315`)

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