Hot copper turnnings can be used as "oxygen getter" for inert gas supplies by slowly passing the gas over the copper tumning at 650 K. 2Cu_((s)) + (1)/(2) O_(2(g)) hArr Cu_(2) O_((s)) : K_(p) = 7.5 xx 10^(10) How many molecules of O_(2)are left in 1 Lof a gas supply after equilibrium has been reached ?

Hot copper turnnings can be used as "oxygen getter" for inert gas supp

1.	Hot copper turnnings can be used as "oxygen getter" for inert gas supplies by slowly passing the gas over the copper tumning at 650 K. 2Cu_((s)) + (1)/(2) O_(2(g)) hArr Cu_(2) O_((s)) : K_(p) = 7.5 xx 10^(10) How many molecules of O_(2)are left in 1 Lof a gas supply after equilibrium has been reached ?
Answer» <P> Solution :PRESSURE is DUE to `O_(2)` in the reaction mixture `K_(P)=(1)/(sqrt(PO_(2)))=7.5 xx 10^(10)` `PO_(2)=(1)/((7.5 xx 10^(10))^(2))=1.778xx 10^(-22)` atm `pv = nRT implies n=(1.778 xx 10^(-22))/(0.0821 xx 650)` `=3.3318 xx 10^(-24)` Molecular = `n xx N` `=3.3318 xx 10^(-24) xx 6 xx 10^(23)=2`