Chapter 12: Problem 78

For the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ at $600 . \mathrm{K},$ the equilibrium constant, $K_{\mathrm{p}},$ is $11.5 .$ Suppose that $2.450 \mathrm{g} \mathrm{PCl}_{5}$ is placed in an evacuated $500 .-\mathrm{mL}$ bulb, which is then heated to $600 .$ K. a. What would be the pressure of $\mathrm{PCl}_{5}$ if it did not dissociate? b. What is the partial pressure of $\mathrm{PCl}_{5}$ at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the percent dissociation of $\mathrm{PCl}_{5}$ at equilibrium?

Short Answer

Expert verified

a) The pressure of $\mathrm{PCl}_{5}$ if it did not dissociate would be approximately $0.02353\, \text{atm}$. b) The partial pressure of $\mathrm{PCl}_{5}$ at equilibrium is approximately $0.00093\, \text{atm}$. c) The total pressure in the bulb at equilibrium is approximately $0.0461\, \text{atm}$. d) The percent dissociation of $\mathrm{PCl}_{5}$ at equilibrium is approximately $96.04 \%$.

Step by step solution

First, we need to calculate the initial moles of $\mathrm{PCl}_{5}$ using the given mass and its molar mass: $2.450 \,\text{g} \mathrm{PCl}_{5} \times (\text{1 mole} / 208.24\, \text{g}) = 0.01176 \,\text{moles} \,\mathrm{PCl}_{5}$ Now we'll use the ideal gas law to calculate the pressure of $\mathrm{PCl}_{5}$ if it did not dissociate. The ideal gas law is $\text{PV}=\text{nRT}$, where $\text{P}$ is the pressure, $\text{V}$ is the volume, $\text{n}$ is moles, $\text{R}$ is the ideal gas constant, and $\text{T}$ is the temperature. Simplify the equation for pressure: $P = \frac{nRT}{V}$. Plug in the given values and solve for the pressure $P=\frac{(0.01176\, \text{moles})(0.0821\,\text{L atm} /(\text{K mol})(600\, \text{K})}{0.5\, \text{L}}$. #Step 2: Set up the reaction table and find expressions relating the partial pressures to the equilibrium constant#

To understand how the system changes when it reaches equilibrium, we'll set up a reaction table, where 'I' stands for Initial, 'C' for Change, and 'E' for Equilibrium: $$ \begin{array}{c|c c c} & \mathrm{PCl}_{5}(g) & \mathrm{PCl}_{3}(g) & \mathrm{Cl}_{2}(g)\\ \hline I & 0.02353\, \text{atm} & 0\, \text{atm} & 0 \, \text{atm} \\ C & -x& +x & +x \\ E & 0.02353 - x & x &x \end{array} $$ The equilibrium constant, $K_\text{p}$, is given by: $K_\text{p} = \frac{[PCl_3][Cl_2]}{[PCl_5]}$. Then the expressions at equilibrium are: $K_p = \frac{x * x}{0.02353-x} = 11.5$ #Step 3: Solving for x and finding the partial pressures at equilibrium#

Next, solve the equation from Step 2 for $x$ which represents the pressures at equilibrium: $11.5 = \frac{x^2}{0.02353-x}$ Rearrange into a quadratic equation: $x^2 + 11.5x - 0.27016 \approx 0$ Solve the quadratic equation for $x$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $x = \frac{-11.5 \pm \sqrt{11.5^2-4(1)(-0.27016)}}{2(1)}$ Taking only the positive root estimate since pressures must be positive, $x \approx 0.0226\, \text{atm}$ Now we can find the partial pressures at equilibrium: $\mathrm{PCl}_{5}(g): 0.02353 - x \approx 0.00093 \, \text{atm}$ $\mathrm{PCl}_{3}(g): x \approx 0.0226 \,\text{atm}$ $\mathrm{Cl}_{2}(g): x \approx 0.0226\, \text{atm}$ #Step 4: Calculate total pressure and percent dissociation at equilibrium#

The total pressure at equilibrium can be found by adding the partial pressures of all the gases: $\text{Total pressure}= 0.00093\, \text{atm} + 0.0226\, \text{atm} + 0.0226\, \text{atm} \approx 0.0461\, \text{atm}$. Next, calculate the percent dissociation of $\mathrm{PCl}_{5}$ at equilibrium: $\text{Percent dissociation} = \frac{0.02353\, \text{atm} - 0.00093\, \text{atm}}{0.02353\, \text{atm}} \times 100 \% \approx 96.04 \%$ Now, we have the answers for each part of the problem: a) The pressure of $\mathrm{PCl}_{5}$ if it did not dissociate would be approximately $0.02353\, \text{atm}$. b) The partial pressure of $\mathrm{PCl}_{5}$ at equilibrium is approximately $0.00093\, \text{atm}$. c) The total pressure in the bulb at equilibrium is approximately $0.0461\, \text{atm}$. d) The percent dissociation of $\mathrm{PCl}_{5}$ at equilibrium is approximately $96.04 \%$.