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Calculate Kc for the following equilibria: 2SO2(g) +O2(g) ====> 2SO3(g) Kp=...
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`2SO_(2(g))+O_(2(g)) harr 2SO_(3(g))`
The relationship of the `K_c` and `K_p` for the above equilibrium is given by;
`k_p = k_c(RT)^(Deltan)`
`Deltan =` (total gas moles of products)-(total gas moles of reactants)
`Deltan = (2-(2+1))`
`R = 8.314 J mol^(-1) K^(-1)`
`T = 500K`
`K_c = (K_p)/(RT)^(Deltan)`
`K_c = (2.5xx10^10)/(500xx8.314)^((2-(2+1)))`
`K_c = 1.039xx10^14M^(-1)`
So the answer is `K_c = 1.039xx10^14M^(-1)`.
According to the value of `K_p` we can assume that it has units of `pa^(-1)`
Posted by jeew-m on July 16, 2013 at 4:46 AM (Answer #1)
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