Question

For the reaction

$\mathit{P}\mathit{C}{\mathit{I}}_{\mathbf{5}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{P}\mathit{C}{\mathit{I}}_{\mathbf{3}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{I}}_{\mathbf{2}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$

at 600. K, the equilibrium constant is 11.5. Suppose that 2.450 g of PCl₅ is placed in an evacuated 500.-mL bulb, which is then heated to 600. K.

a. What would the pressure of PCl₅ be if it did not dissociate?

b. What is the partial pressure of PCl₅ at equilibrium?

c. What is the total pressure in the bulb at equilibrium?

d. What is the degree of dissociation of PCl₅ at equilibrium?

Short answer

a. The pressure of PCI₅ would be 1.16atm .

b.The partial pressure of PCI₅ at equilibrium is 0.10atm .

c.The total pressure in the bulb at equilibrium is 2.22atm .

d.The degree of dissociation of PCI₅ at equilibrium is 91.4%.

Step-by-step solution

Find the initial pressure of PCl5.

The initial pressure of PCl₅ can be calculated by the ideal gas law as follows:

$P=\frac{nRT}{V}$ …(1)

Here,

n is the number of moles.

P is the pressure.

V is the volume.

T is the temperature.

R is the gas constant, 0.08206Latm/molK .

$\begin{array}{rcl}{P}_{PC{I}_3}& =& \frac{{\displaystyle \frac{2.450g}{208.22g/mol}}\times 0.08206Latm/Kmol\times 600.K}{0.500L}\\ & =& 1.16atm\end{array}$

The initial pressure of PCl₅ would be 1.16atm.

Calculate the partial pressure of PCl5 at equilibrium.

The equilibrium constant K_p expression for this reaction is written below:

$K_p=\frac{\left(P_{PC{I}_3}\right)\left(P_{C{I}_2}\right)}{P_{PC{I}_5}}$ …(2)

Construct the ICE table:

$$\begin{gathered} PC{I}_5\rightleftharpoons PC{I}_3+C{I}_2 \\ I\left(atm\right)1.1600 \\ C\left(atm\right)-x+x+x \\ -------------------- \\ E\left(atm\right)1.16-xxx \end{gathered}$$

Substitute these values and the value of Kp into Equation (2).

$11.5=\frac{x^2}{1.16-x}$

Solve for x

$\begin{array}{rcl}x& =& 1.06\\ & & \\ P_{PC{I}_3}& =& 1.16-X\\ & =& 1.16-1.06\\ & =& 0.10atm\\ & & \\ P_{PC{I}_3}& =& x\\ & =& 1.06atm\\ & & \\ P_{C{I}_2}& =& x\\ & =& 1.06atm\end{array}$

Therefore, the partial pressure of PCl₅ at equilibrium is 0.10atm.

Find the total pressure in the bulb at equilibrium.

The calculated partial pressures of PCI₅(g) , PCI₃(g) and CI₂(g) are 0.10atm, 1.06atm, and 1.06atm , respectively.

The total pressure in the bulb at equilibrium can be determined as follows:

$\begin{array}{rcl}{P}_{total}& =& P_{PC{I}_5}+P_{PC{I}_3}+P_{C{I}_2}\\ & =& 0.10atm+1.06atm+1.06atm\\ & =& 2.222atm\end{array}$

Hence, the total pressure in the bulb at equilibrium is 2.22atm .

Find the degree of dissociation of PCl5 at equilibrium.

The degree of dissociation of PCl₅ at equilibrium can be found as follows:

$\begin{array}{rcl}Percentdissociation& =& \frac{x}{initialpressure}\times 100\%\\ & =& \frac{1.06atm}{1.16atm}\times 100\%\\ & =& 91.4\%\end{array}$

Therefore, the degree of dissociation of PCl₅ at equilibrium is 91.4%.