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

`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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