N2+O2<---->2NO. The Mass Law Equilibrium Constant is1.00X10^-30 at 25 degrees Ce

Question

N2+O2<---->2NO. The Mass Law Equilibrium Constant is1.00X10^-30 at 25 degrees Celcius. Starting with a concentration ofN2 at 0.030 moles in a 3L flask and O2 the same, calculate theequilibrium concentration of each species at 25 degreesCelcius. This problem is not in any textbook, but on a worksheet Ihave. N2+O2<---->2NO. The Mass Law Equilibrium Constant is1.00X10^-30 at 25 degrees Celcius. Starting with a concentration ofN2 at 0.030 moles in a 3L flask and O2 the same, calculate theequilibrium concentration of each species at 25 degreesCelcius. This problem is not in any textbook, but on a worksheet Ihave.

Explanation / Answer

For the given reaction,                 N2+O2 <----> 2NO We have - Initial concentration of the N2 = 0.030/3L = 0.01M Concentration of O2 = 0.01 M Thus -                       N2 +   O2 <----> 2NO            I(M)   0.01   0.01              0             C       - x      -x                +2x            Eq      0.01-x   0.01-x         2x therefore -                                   1.00*10-30 = 2x^2 / (0.01-x)^2 By solving it we get                                            x = 7.017*10-18 Thus atequilibrium          [NO] =  7.017*10-18, [O2] = [N2] =0.01 -  7.017*10-18 = 0.00999M Thus atequilibrium          [NO] =  7.017*10-18, [O2] = [N2] =0.01 -  7.017*10-18 = 0.00999M

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