Problem 80 - Chemistry The Central Science In Si Units

Question

For the equilibrium $$ \mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g) $$ \(K_{p}=5.27\) at \(60^{\circ} \mathrm{C} .(\mathbf{a})\) Calculate \(K_{c} .(\mathbf{b})\) After \(3.00 \mathrm{~g}\) of solid \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) is added to a closed 1.500-L vessel at \(60^{\circ} \mathrm{C}\), the vessel is charged with \(0.0500 \mathrm{~g}\) of \(\mathrm{BCl}_{3}(g)\). What is the equilibrium concentration of \(\mathrm{PH}_{3}\) ?

Step-by-Step Solution

Answer

(a) Kc is 0.0194; (b) Equilibrium concentration of PH3 is 0.0681 M.

Step 1: Relationship between Kp and Kc

The relationship between the equilibrium constants Kp and Kc is given by the formula \[K_p = K_c (RT)^{\Delta n}\]where \(R\) is the gas constant \(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\), \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas. For this reaction, \(\Delta n = 2 - 0 = 2\).

Step 2: Convert Temperature to Kelvin

The given temperature is 60°C. To convert this to Kelvin, use the formula:\[ T(K) = T(°C) + 273.15\]Therefore, \(T = 60 + 273.15 = 333.15 \, K\).

Step 3: Calculate Kc

Now, substitute the known values into the relationship equation:\[5.27 = K_c (0.0821 \times 333.15)^2\]Solve for \(K_c\) to find:\[K_c = \frac{5.27}{(0.0821 \times 333.15)^2} \approx 0.0194\].

Step 4: Determining Initial Concentrations

The initial concentration of \(\mathrm{BCl}_3\) is calculated by converting grams to moles:\[0.0500 \, \text{g} \times \frac{1 \, \text{mol}}{117.16 \, \text{g mol}^{-1}} = 4.27 \times 10^{-4} \, \text{mol}\]The concentration is:\[\frac{4.27 \times 10^{-4}}{1.500} = 2.85 \times 10^{-4} \, \text{M}\]There is initially no \(\mathrm{PH}_3\) in the gas phase.

Step 5: Establish ICE Table

Construct an ICE table (Initial, Change, Equilibrium) for concentrations:- Initial: \([\mathrm{BCl}_3] = 2.85 \times 10^{-4} \, \text{M}\), \([\mathrm{PH}_3] = 0\)- Change: \(-x\) for \(\mathrm{BCl}_3\), \(+x\) for \(\mathrm{PH}_3\)- Equilibrium: \([\mathrm{BCl}_3] = 2.85 \times 10^{-4} - x\), \([\mathrm{PH}_3] = x\)

Step 6: Apply Kc to ICE Table

The equilibrium expression using \(K_c\) is:\[K_c = \frac{[\mathrm{PH}_3][\mathrm{BCl}_3]}{(1)\text{ (solid doesn't appear in K-expression)}} \]Using the ICE table:\[0.0194 = \frac{x (2.85 \times 10^{-4} - x)}{1}\]Assume \(x\) is very small compared to \(2.85 \times 10^{-4}\), hence:\[0.0194 \approx x (2.85 \times 10^{-4})\]

Step 7: Solve for x

Solving \[0.0194 = x (2.85 \times 10^{-4})\],\[x = \frac{0.0194}{2.85 \times 10^{-4}} \approx 6.81 \times 10^{-2} \, M\].This gives the equilibrium concentration of \(\mathrm{PH}_3\) as \(6.81 \times 10^{-2} \, M\).

Key Concepts

Equilibrium Constant

The equilibrium constant is a crucial concept in chemical equilibrium. It helps predict the direction of a chemical reaction and its extent. There are two forms of equilibrium constants: \(K_c\) and \(K_p\).

• \(K_c\) uses concentrations of substances in molarity (\(M\)) at equilibrium.
• \(K_p\) deals with partial pressures of gases involved in the reaction.

The relationship between these two constants is particularly important: \[K_p = K_c (RT)^{\Delta n}\]where \(R\) is the gas constant with a value of \(0.0821\, \text{L atm K}^{-1} \text{mol}^{-1}\), \(T\) is the temperature in Kelvin, and \(\Delta n\) represents the change in moles of gas.
To convert the temperature from Celsius to Kelvin, add \(273.15\) to the Celsius value. This conversion is essential because gas laws and related calculations usually use Kelvin as the temperature unit.
Once temperature is in Kelvin, use the equation for \(K_p\) and \(K_c\) to find the required equilibrium constant. This conversion allows calculations dealing with effects under different conditions in reactions involving gases.

ICE Table

An ICE table is a valuable tool in chemical equilibrium that organizes information about the initial concentrations, changes, and equilibrium concentrations of reactants and products.
Use it to systematically determine unknown values in calculations involving equilibrium. Here's how it works:

• Initial: Write down initial concentrations (or pressures) of all reactants and products. If a substance is not present initially, use zero.
• Change: Represent changes in concentrations with \(x\). For a reaction \(\text{A} \rightleftharpoons \text{B} + \text{C}\), if \(A\) decreases by \(x\), then \(B\) and \(C\) increase by \(x\).
• Equilibrium: Calculate equilibrium concentrations by adding initial values to changes: \([A]_0 - x\), \([B]_0 + x\), \([C]_0 + x\).

This systematic approach provides a clear view of how concentrations shift from start to equilibrium, helping solve for unknowns using equilibrium expressions.
Keep in mind that if \(x\) is much smaller compared to initial concentrations, it can be neglected in calculations for simplification, improving ease and clarity in problem-solving.

Gas Law Calculations

Gas laws play an integral role when dealing with gases in chemical equilibrium problems. The ideal gas law equation is used frequently:\[PV = nRT\]where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature in Kelvin.
Gas law calculations help relate the amount of gas to its partial pressure, which can be expressed using: - \(P = \frac{nRT}{V}\)This equation will let you calculate the pressure of a gas if you know its volume and temperature, which is vital information for determining equilibrium positions involving gases.
Use these calculations to determine partial pressures or convert between units. These values feed into the equilibrium constant calculations (\(K_p\) and \(K_c\)), linking thermodynamic aspects to kinetic expressions of chemical reactions.
In problems involving gases, always remember:

• Convert temperatures to Kelvin.
• Use consistent units, like liters for volume and atm for pressure.
• Understand how changes in pressure can shift equilibrium, following Le Chatelier's principle.

These principles ensure accurate and reliable conclusions in solving equilibrium scenarios.