For the following reaction, Kp = 6.5x10^4 at 308 K: 2NO(g) + Cl2(g) ===> 2NOCl(g)

At equilibrium, PNO= 0.35 atm and PCl2 = 0.10 atm. What is the equilibrium partial pressure of NOCl(g)?

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`2NO_((g))+Cl_(2(g)) harr 2NOCl_((g))`

`K_p = (P_(NOCl))^2/((P_(Cl_2))(P_(NO))^2)`

According to the given data;

`K_p = 6.5x10^4pa^(-1) = 0.6415atm^(-1)`

`P_(NO) = 0.35 atm`

`P_(Cl_2) = 0.10 atm`

`0.6415= (P_(NOCl))^2/(0.1xx0.35^2)`

`P_(NOCl) = 0.963atm`

*So the partial pressure of NOCl is 0.963atm.*

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