Question

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C If you place 1.1 g of the complex in a 69.5 ml flask at 125 °C, what is the total pressure in the flask at equilibrium? What are the partial pressures of the Lewis acid, the Lewis base, and the complex? P(BF) - atm P((CH),0) - atm P((CH)20-BF) P(total)- am atm

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Answer 1

The lewis substance which is e acid. deficient and accept pair fe is called The substance which donate e pair called lewis base. = BF; (g) + (043)0 (9) (439)0 —>BF3 BF₃ is lewis acid- (CH3) is lewis base Calculating initial Pressure of complex $ven weight 1 согорlеx = 1 3 - Temprature=125² T(K)ste +2 73.15 = 125+273.15 = 398.15k Volume = 69.5 mL im = 0.0012 69.5mL = 0.06952 molar mas a complex = 113.87 mol mol = given mass molar mass lolg 50.00 966 mol 73.81 fmolt PV=nRT v= volume p= nRT p= to n= number of mol R = gas constanz T = Temp.

Pa 0.00966 1901 x 0.0821 x ahm X 398.156 neto 0.06954 P = 4.54 atom (CH3) 0-BF3 (g) 4.54 = BF3(g) + (CH3), 0 (9) Initial (atm) -Þ þ change b At 4.54-1 epubางๆ Kp = (BF) (CM3),0) (CH3O+BF3) (PDCP) 0.17 = 4.54-b 0.712 -0.17 = Þ2 +0.175 -0.772=0 Compare this equation ax²+bx+c=o. a=l b= 0.116=-0.772 pe= -bl Jb2_uac 20 = -0.17 + 1T0.1792 -4*-0.772 - 2*1

-0.17 + Jo.0289 + 3.088 -0.17+ 3.1169 2 0.80 so PCBF3) = þ (CM3),0 = b = 0.80 am b = 0.80 am 4.544- b = 4.54-0.80 = 3-74 PCM3.0 - BF, = 0.80) am = 5.34 am Total Pressne=13.74 +0.80+