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Reaction between N(2) and O(2-) takes pl...

Reaction between `N_(2)` and `O_(2-)` takes place as follows :
`2N_(2)(g) + O_(2)(g) hArr 2N_(2)O (g)`
If a mixture of `0.482` mol `N_(2)` and `0.933` mol of `O_(2)` is placed in a `10 L` reaction vessel and allowed to form `N_(2)O` at a temperature for which `K_(c) = 2.0 xx 10^(-37)`, determine the composition of equilibrium mixutre.

Text Solution

Verified by Experts

As `K = 2.0 xx 10^(-37)` is very small, this means that the amount of `N_(2) "and" O_(2)` reacted (x) is very small, Hence, at equilibrium, we have `[N_(2)] = 0.0482 molL^(-1), [O_(2)] = 0.0933 molL^(-1), [NO_(2)] = 0.1x`
`K_(c) = (0.1x)^(2)/((0.0482)^(2)(0.0933)) = 2.0 xx 10^(-37)`
On solving, this gives `x = 6.6 xx 10^(-20), [N_(2)O] = 0.1x = 6.6 xx 10^(-21) molL^(-1)`
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