at 200 oC Kc = 1.4 x 10-10 for the reaction N2O(g) + NO2(g) <-> 3NO(g)....

at 200 ^oC K_c = 1.4 x 10^-10 for the reaction N₂O(g) + NO₂(g) <-> 3NO(g). If 300 ml of NO measured at 800 torr and 25 ^oC is placed in a 4.00 L container. what will be the N_²O and NO molar concentrations at equilibrium? What will be the total pressure of the mixture (in torr) at equilibrium at 25 ^oC?

14.98 Jespersen 7th Ed.

Solutions

Expert Solution

Moles of No can be calculated from ideal gas law equation from

PV= nRT, n= PV/RT, P= 800Torr= 800/760 atm =1.052 atm, V= 300ml= 300/1000L=0.3L, T= 25+273= 298K, R= 0.0821 L.atm/mole.K, n= 1.052*0.3/(0.0821*298)=0.0129 moles

Molar concentration of NO= 0.0129/4=0.003225M

The reaction is 3NOßà N2O+ NO2, Kc’= 1/KC= 10¹⁰/1.4= 7.14*10⁹

since there is neither N2O nor NO2, the reaction begins only with the decomposition of NO.

Let x= concentration of N2O at equilibrium.

Hence at equilibrium [N2O]= [NO2]=x and [NO]= 0.003225-3x

Hence KC’ = 7.14*10⁹ = x²/(0.003225-3x)³

When solved using excel, x= 0.0010731855

At Equilibrium [N2O]= [NO2] =0.0010731855 and [NO] =5.4435*10-6 M

Moles at equilibrium

[N2O]= 4*0.0010731855 = [N2O] = 0.00429272

[NO]= 4*5.4435*10^-6 =2.1774*10-5 moles

Total moles= 2*0.00429272+2.1774*10^-5 = 0.00861moles

From gas law, P= nRT/V= 0.00861*0.0821*298/4= 0.053 atm =0.053*760 torr= 40.28 torr

queen_honey_blossom answered 1 year ago

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