The normal boiling temperature of benzene, C_6H_6, is 80.25 degree C, whilst the

Question

The normal boiling temperature of benzene, C_6H_6, is 80.25 degree C, whilst the vapor pressure at a temperature of 60.6degree C is 53.0 kpa Determine the enthalpy of vaporization of benzene. when dissolving 2.42 g of a nonvolatile compound in 7.8 g of benzene at 60.6degree C, the vapor pressure fell to 50.35 kpa. Calculate the molar mass of the compound. Argon, the third most abundant component in air (1%), is produced from a complex process of compression, cooling and distillation, Assuming the following separation process starting with 1000 moles of air with the given composition, 1000 mol Air(780 mol N and AG for the separation process; (b) Calculate AS for the process. Reaction A(g) + B (s) 2C (g) at 400 K and 100 bar has AM = 3.3256 kJ mol^-1 and nablaS = 8.314 J mol*' K'' (a) Determine the equilibrium constant of the reaction at 400 K; (b) what assumption do you have to make to determine the equilibrium constant at 800 K: (c) determent the equilibrium constant at 800 K and 50 bar. Limestone, CaCO, decomposes upon heating to make CaO and CO2 by the following reaction: CaC03(5) = Ca0(^) + C02(g) Given the following thermodynamic properties: AiH^(CaC03)= -1206.9 kJ/mol, AfG®(CaC03)= -1128.8 kJ/moI; AfH^(CaO)= -635.09 kJ/mol, AfG'^(CaO)= -604.0 kJ/mol; AfHC02)= -393.5 IkJ/mol, AiG degree (C02)= -394.36 kJ/mol. (a) Determine the pressure of CO2 at room temperature; (b) Assuming ArH is independent of temperature, calculate the temperature at which the pressure of CO2 is 1 bar. The standard cell potential for the Daniel cell Zn(s)IZnS04(aq)l|CuS04(aq)lCufs) is+1.102 V, Write the cell reaction and determine its equilibrium constant K at 25 degree C. Determine the cell potential at 25 degree C if the concentration of zinc sulfate, ZnS04 is 0.050 mol dm- and the concentration of copper sulfate, CuS04, is 0.200 mol dm^-3

Explanation / Answer

Apply Clasius Clapeyron equation

ln(P2/P1) = H/R*(1/T1- 1/T2)

substitute known data

ln(53/101.3) = H/(8.314)*(1/(80.25+273) - 1/(60.6+273))

H = ln(53/101.3) * 8.314 / (1/(80.25+273) - 1/(60.6+273))

H = 32,299.28 J/mol = 32.29 kJ/mol

b)

2.42 g of compound

m = 7.8 g of benzene

T = 60.6 , vapor pressure drops to 50.35 kPa

dP = xsolute * Psolvent

(53-50.35) = xsolute * 53

xsolute = (53-50.35) /53 = 0.05

mol of benzene = mass/MW = 7.8/78 = 0.1 mol

so

0.05 = mol of x / ( mol of x + 0.1)

0.05*x + 0.05*0.1 = x

0.95x = 0.005

x = (0.005)/(0.95) = 0.0052631 mol

MW = mass/mol = 2.42/0.0052631 = 459.805 g of sample

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