Reaction between `N_2` and `O_(2-)` takes place as follows : `2N_(2(g)) + O_(2(g)) hArr 2N_2O_((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 `N2O` at a temperature for which `K_c = 2.0 xx 10^(-37)`, determine the composition of equilibrium mixture.

Explanation

`{:("Reaction:",2N_(2(g)) + , O_(2(g)) hArr , 2N_2O_((g))),("Initial mol:",0.482,0.933,0),("Change in reaction:",-2x,-x,+2x),("Mol at equilibrium:",(0.482-2x),0.933-x,2x),("Equilibrium mol L"^(-1), "0.482-2x"/10,"0.933-x"/10,(2x)/10):}`
`[N_2]=(0.482-2x)/10 = 0.482/10 "mol L"^(-1) = 0.0482 "mol L"^(-1)`
`[O_2]= (0.933-x)/10 = 0.933/10 "mol L"^(-1) = 0.0933 "mol L"^(-1)`
`[N_2O]=(2x)/10 = 0.2x "mol L"^(-1)`
`K_c=[N_2O]^2/([N_2]^2[O_2])=(0.2x)^2/((0.0482)^2(0.0933))`
`therefore 2.0xx10^(-37)=(0.04x^2)/((0.0482)^2(0.0933))`
`therefore 0.04x^2=2.0xx10^(-38)xx(0.0482)^2(0.0933)`
`therefore x=sqrt((20xx10^(-38)xx(0.0482)^2(0.0933))/(0.04))`
`therefore x=sqrt(0.1084xx10^(-38))= 0.3292xx10^(-19)`
Now, `[N_2O]=2x=2(0.3292xx10^(-19))`
`=0.6584xx10^(-19)=6.584xx10^(-20) "mol L"^(-1)`
`[N_2]=0.482-2x=0.482-0.6584xx10^(-19) approx 0.482`
`[O_2]= 0.0933-x=0.0933-0.6584xx10^(-19) approx` 0.0933