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.

Explanation

`2N_2(g)+O_2(g) hArr 2N_2O(g)`
`K_c=([N_2O]^2)/([N_2]^2 [O_2])`
Let us suppose that x mol of `O_2` has reacted with 2x mol of `N_2` to form 2x mol of `N_2O`. Then at equilibrium the concentrations are :
`N_2`=0.482 -2x, `O_2`=0.933-x and `N_2O` =2x
Since the volume of reaction vessel is 10.0 L , so the molar concentrations at equilibrium are :
`[N_2]=(0.482-2x)/(10.0)=0.0482-0.2x "mol L"^(-1)`
`[O_2]=(0.933-x)/10.0 =0.0933-0.1 "mol L"^(-1)`
`[N_2O]=(2x)/10.0 =0.2x "mol L"^(-1)`
`therefore [K_c]=(0.2x)^2/((0.0482-0.2x)^2(0.0933-0.1x))`
Since `K_c` is very small `(2.0xx10^(-37))` , we can assume x to be very small so that
0.0482-0.2x `approx` 0.0482 and 0.0933-0.1x `approx` 0.0933
`K_c=(0.04x^2)/((0.0482)^2(0.0933))`
or `0.04x^2=(0.0482)^2 (0.0933)xx(2.0xx10^(-37))`
or `0.04x^2=4.34xx10^(-41)`
`x^2=(4.34xx10^(-41))/0.04=1.085xx10^(-39)`
`therefore x=3.3xx10^(-20)`
Thus , equilibrium concentrations are :
`[N_2]`=0.0482-0.2x=0.0482
`[O_2]` =0.0933-0.1x=0.0933
`[N_2O]` = 0.2x=`0.2xx3.3xx10^(-20)=6.6xx10^(-21)`