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You are watching: 2so2(g)+o2(g)⇌2so3(g)


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`2SO_(2(g))+O_(2(g)) harr 2SO_(3(g))`

The connection of the `K_c` and `K_p` because that the over equilibrium is offered by;

`k_p = k_c(RT)^(Deltan)`

`Deltan =` (total gas mole 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...


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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 provided by;

`k_p = k_c(RT)^(Deltan)`

 

`Deltan =` (total gas mole of products)-(total gas mole 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 prize is `K_c = 1.039xx10^14M^(-1)`.

See more: How Does Impaired Driving Rank As A Highway Safety Problem ?

 

 

Note:

According to the worth of `K_p` we deserve to assume the it has units the `pa^(-1)`


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