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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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Reaction between nitrogen and oxygen takes place as following: 2N_(2(g))+O_(2)hArr2N_(2)O_((g)) If a mixture of 0.482 "mole" N_(2) and 0.933 "mole" of O_(2) is placed in a reaction vessel of volume 10 litre and allowed to form N_(2)O at a temperature for which K_(c)=2.0xx10^(-37)litre mol^(-1) . Determine the composition of equilibrium mixture.

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Knowledge Check

  • For a reaction, 2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g) rate of reaction is:

    A
    Rate `= - (d[N_(2)O_(5)])/(dt) = - (1)/(4) (d[NO_(2)])/(dt) = (1)/(2)(d[O_(2)])/(dt)`
    B
    Rate `= - (1)/(2) (d[n_(2)O_(5)])/(dt) = (1)/(4)(d[NO_(2)])/(dt) = (d[O_(2)])/(2)`
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    D
    Rate `= - (1)/(2) (d[N_(2)O_(5)])/(dt) = (1)/(2) (d[NO_(2)])/(dt) = (1)/(2) (d[O_(2)])/(dt)`

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