Calculate Kc for the following equilibria: 2SO2(g) +O2(g) ====> 2SO3(g) Kp= 2.5x10^10 at 500K



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Posted on (Answer #1)

`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)`

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