Equilibrium 1.00-2x Thus at equilibrium, 0-K,=a00-2x x(x) P'No x2=K.1.00-2x)2 x2=K.1.00-4x+4x2) x2-4Kx2+4Kx-K。=0 1-4K.)x2+4K,x-K。=0 a=1-4K。=1-41.78×10)=-7.12×10 b=4K。=7.12×10 c=-K。=-1.78×10 x-b±v62-4ac -7.12×10±√7.12×10)2-4(-7.12×10-1.78×10) 2a 2(-7.12×10) =-7.12x102±0 -14.24×10 =0.5 Thus,PNo=1.00-2x=0 Po2 PN2 =x=0.5 atm (This indicates essentially complete dissociation of the NO to O2 and N2.which could be predicted from the very large value of Kp for this reaction.) 4 Use the ideal gas law to calculate the pressure(atm)exerted by 1.00 mol of N in a 1.00 Lcontainer at 25C p= nRT 1.00 mol(0.082L atm K-'mol-)(25+273)K 1.00L -24.4atm 5.Iridium(Ir)has a face centred cubic unit cell with an edge length of 383.3 pm.Calculate the density of solid Iridium(g/cm). The unit cell contains the equivalent of 4 atoms: 6Equilibrium 1.00-2x x x Thus at equilibrium, N O 2 2 2 2 p NO 2 2 p 2 2 p 2 2 p pp 2 p pp 7 7 p 7 p 7 p 2 7 7 2 p p x(x) K p (1.00 2x) x K (1.00 2x) x K (1.00 4x 4x ) x 4K x 4K x K 0 (1 4K )x 4K x K 0 a 1 4K 1 4(1.78 10 ) 7.12 10 b 4K 7.12 10 c K 1.78 10 b b 4ac 7.12 10 (7.12 10 ) 4( 7.1 x 2a = = − = − = −+ − + −= − + −= = − = − × =− × ==× =− =− × −± − − × ± × −− = = 7 7 7 7 7 2 10 )( 1.78 10 ) 2( 7.12 10 ) 7.12 10 0 14.24 10 0.5 × −× − × − ×± = − × = Thus, pNO = 1.00 – 2x = 0 pO2 = pN2 = x = 0.5 atm (This indicates essentially complete dissociation of the NO to O2 and N2, which could be predicted from the very large value of Kp for this reaction.) 4 Use the ideal gas law to calculate the pressure (atm) exerted by 1.00 mol of N2(g) in a 1.00 L container at 25o C. 1 1 nRT 1.00 mol(0.082L atm K mol )(25 273)K p V 1.00 L 24.4 atm − − + = = = 5. Iridium (Ir) has a face centred cubic unit cell with an edge length of 383.3 pm. Calculate the density of solid Iridium (g/cm3 ). The unit cell contains the equivalent of 4 atoms: 6