Question

The degree of dissociation is 0.4 at 400k and 1atm for gaseous reaction

Answer 1

At 1 atm and 400 K, there is a mixture of PCl5, PCl3 and Cl2 with number of moles of 0.600, 0.400, and 0.400, respectively.  

The density of the mixture is the sum of the densities of the three gases can be shown as:

$Total density of gas mixture = d_P_C_l_5 + d_P_C_l_3 + d_C_l_2$

Density can be calculated as:

$D = PM/RT$

D=density of a gas,  

P=pressure of gas  

$Total moles = n_P_C_l_5 + n_P_C_l_3 + n_C_l_2$

$= 0.600 mol + 0.400 mol + 0.400 mol = 1.40 mol$

$Mole fraction (X) of each gas$

$X(PCl_5) = 0.6 mol / 1.40 mol = 0.4286$

$X(PCl_3) = 0.4 mol / 1.40 mol = 0.2857$

$X(Cl_2) = 0.4 mol / 1.40 mol = 0.2857$

Partial pressures of each gas: Pn = Xn(Pt)

$P(PCl5) = X(PCl5) x 1.00 atm = 0.4286 atm$

$P(PCl3) = X(PCl3) x 1.00 atm = 0.2857 atm$

$P(Cl2) = X(Cl2 x 1.00 atm = 0.2857atm$

$d = PM/RT$

$d(PCl5) = 0.4287 x 208.5g/mol / 0.0821 Latm/molK / 400K = 2.721 g/L$

$d(PCl3) = 0.2857 x 137.5g/mol / 0.0821 Latm/molK / 400K = 1.196 g/$

$d(Cl2) = 0.2857 x 71.0g/mol / 0.0821 Latm/molK / 400K = 0.618 g/$

Thus, Total density of gas mixture = 2.721g/L + 1.196g/L + 0.618g/L = 4.53 g/L

Thus, Total density of gas mixture is 4.53 g /L