# Answer on Physical Chemistry Question for nidhi mehra

Question #21243

the reaction N2+O2=2NO2 contributes to air pollution. whenever a fuel is burnt in air at high temp. At 1500 K equilibrium constant K for it is 1.0*10^-5. suppose in a case [N2]=0.80 mol per litre and [O2]= 0.20 mol per litre before any reaction occurs. Calculate the equillibrium concerntrations of the reactants and the products after the mixture has been heated to 1500 K.

Expert's answer

Equilibrium constant can be calculated by next equation:

K = [products]/[reactants]

K = [NO

Concentration after reaction would be: (where x is that part

of reactants that would be converted in product)

0.8-x 0.2-x 2x

N

In the moment at equilibrium equilibrium constant would be:

K = [2x]

1.0*10^-5 * (0.8-x)*( 0.2-x) = 4x

1.0*10^-5 * (0.6-0.8x-0.2x+x^2) = 4x

6.0*10^-6 - 6.0*10^-6 x + 1.0*10^-5x^2 - 4x^2 = 0

x = 0.001224 M so concentration of product NO

concentration of O

K = [products]/[reactants]

K = [NO

_{2}]^{2}/[O_{2}][N_{2}]Concentration after reaction would be: (where x is that part

of reactants that would be converted in product)

0.8-x 0.2-x 2x

N

_{2}+ O_{2}= 2NO_{2}In the moment at equilibrium equilibrium constant would be:

K = [2x]

^{2}/[0.8-x][0.2-x]1.0*10^-5 * (0.8-x)*( 0.2-x) = 4x

^{2}1.0*10^-5 * (0.6-0.8x-0.2x+x^2) = 4x

^{2}6.0*10^-6 - 6.0*10^-6 x + 1.0*10^-5x^2 - 4x^2 = 0

x = 0.001224 M so concentration of product NO

_{2}m = 2*0.001224 = 0.002448 Mconcentration of O

_{2}= 0.2-0.001224 = 0.198776 and concentration of N_{2}= 0.798776 MNeed a fast expert's response?

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