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For the following reaction, Kp = 6.5x10^4 at 308 K: 2NO(g) + Cl2(g) ===> 2NOCl(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`
`P_(NOCl) = 0.963atm`
So the partial pressure of NOCl is 0.963atm.
Posted by jeew-m on July 16, 2013 at 5:06 AM (Answer #1)
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