For the chemical system shown below, it was found that[N 2 ] = [O 2 ] = 0.70 mol

Question

For the chemical system shown below, it was found that[N2] = [O2] = 0.70 mol/L and [NO] =0.22mol/L, at equilibrium. Suppose that enough N2 wasadded to increase its concentration temporarily to 1.00 equilibriumconcentrations of N2, O2, and NO? N2 + O2(g) <-> 2NO(g) Thanks! For the chemical system shown below, it was found that[N2] = [O2] = 0.70 mol/L and [NO] =0.22mol/L, at equilibrium. Suppose that enough N2 wasadded to increase its concentration temporarily to 1.00 equilibriumconcentrations of N2, O2, and NO? N2 + O2(g) <-> 2NO(g) Thanks!

Explanation / Answer

   The given reaction                                     N2 (g) +O2  (g) ------> 2NO(g)                             But at the euilibrium the concentration of N2 and O2 areequal.                                          [N2]= [O2] =0.70 mol/L                                    and   [NO] =0.22mol/L                                                 Rate constant Kc = [NO]2 / [N2][O2]                                                       =(0.22)2 /(0.70) (0.70)                                                       = 0.0484/0.49                                                       = 0.0987                           If N2 concentration as increased to 1.00 then there is nochange the other two concentrationof     NO,O2       But the final rate is changed at equilibrium.                                                   

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