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the degree of dissociation is 0.4 at 400K and 1 atm for PCl5. Assuming ideal gas behaviour for all gases, calculate the density of equilibrium mixture at 400K and 1 atm.
the degree of dissociation is 0.4 at 400K and 1 atm for PCl5. Assuming ideal gas behaviour for all gases, calculate the density of equilibrium mixture at 400K and 1 atm.

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3 years ago

```							PCl5(g)  PCl3(g) + Cl2(g) We are told that at 1.00 atm of pressure there is a mixture of PCl5, PCl3 and Cl2 with the relative number of moles of 0.600, 0.400, and 0.400, respectively. This comes from the "degree of dissociation." The density of the mixture will be the sum of the densities of the three gases. We can sum the densities because each gas occupies the same volume. Since masses are summative, and volume is fixed, the densities are summative. Density is calculated from this equation: D = PM/RT ..... D=density of a gas, P=partial pressure of that gas Therefore, we must compute the partial pressure of each gas, and the density of each gas. We will use the concept of "mole fraction." Mole fraction, X, is the moles of a component of the mixture divided by the total moles. The partial pressure of the first gas in the mixture is related to the mole fraction: P1 = X1Pt, where Pt is the total pressure, and X1 is the mole fraction of the first component. The partial pressure of each gas can then be calculated in the same way. Total moles = 0.600 mol + 0.400 mol + 0.400 mol = 1.40 mol Mole fraction (X) of each gas X(PCl5) = 0.6 mol / 1.40 mol = 0.4286 X(PCl3) = 0.4 mol / 1.40 mol = 0.2857 X(Cl2) = 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 Density calculations: 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/L D(Cl2) = 0.2857 x 71.0g/mol / 0.0821 Latm/molK / 400K = 0.618 g/L Total density of gas mixture = 2.721g/L + 1.196g/L + 0.618g/L = 4.53 g/L ...... to three sig figs
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3 years ago
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