Question

For the equilibrium $$\mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g)$$ (b) After \(3.00 \mathrm{~g}\) of solid$$K_{p}=0.052 \text { at } 60^{\circ} \mathrm{C} \text { . (a) Calculate } K_{c}$$ \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) is added to a closed \(1.500-\mathrm{L}\) 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 \(\mathrm{PH}_{3} ?\)

Short answer

The equilibrium concentration of PH3 is approximately 1.03 x 10^{-4} M.

Step-by-step solution

Write the expression for Kp in terms of Kc

Recall that Kp and Kc are related by the equation: \[K_p = K_c(RT)^{\Delta n}\] where R is the gas constant (0.0821 Latm/molK), T is the temperature in Kelvin, and Δn is the difference in the moles of the gaseous products and reactants.

Determine the value of Δn

In the balanced equilibrium reaction, \[\mathrm{PH}_{3}\mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g)\] Δn = moles of gaseous products - moles of gaseous reactants = (1 + 1) - 0 = 2.

Convert the temperature to Kelvin

Given that the temperature is 60°C, we will first need to convert it to Kelvin: \[60^{\circ} \mathrm{C} = 60 + 273.15 = 333.15 \mathrm{K}\]

Calculate Kc

Now use the given value of Kp and the calculated value of Δn to find Kc: \[K_c = \frac{K_p}{(RT)^{\Delta n}}\] \[K_c = \frac{0.052}{(0.0821 \times 333.15)^2}\] \[K_c \approx 1.87 \times 10^{-6}\] (b) Finding the equilibrium concentration of PH3

Calculate the moles of the reactant and product

Given that 3.00 g of solid PH3BCl3 is added, we need to convert it to moles: \[\text{moles of PH}_{3}\text{BCl}_{3} = \frac{3.00 \text{ g}}{269.65 \text{ g/mol}} \approx 0.0111 \text{ mol}\] Similarly, as 0.0500 g of BCl3(g) is already present in the vessel, we need to convert it to moles: \[\text{moles of BCl}_{3(g)} = \frac{0.0500 \text{ g}}{117.16 \text{ g/mol}} \approx 4.27 \times 10^{-4}\text{ mol}\]

Set up an ICE table

Initially, we have 0 mol of PH3 in the reaction mixture. Now, let's set up an ICE table (Initial, Change, Equilibrium) to find the equilibrium concentrations of the reactants and products. | | PH3BCl3(s) | PH3(g) | BCl3(g) | |---|-----------------|---------------------|---------------------| | I | 0.0111 mol | 0 mol | 4.27 x 10^-4 mol | | C | -x mol | +x mol | +x mol | | E | 0.0111-x mol | x mol | (4.27 x 10^-4+x) mol | Where x is the mol change of PH3 and BCl3.

Write the expression for Kc

Using the equilibrium concentrations from the ICE table, the expression for Kc becomes: \[K_c = \frac{[\mathrm{PH}_3][\mathrm{BCl}_3]}{[\mathrm{PH}_3\mathrm{BCl}_3]}\]

Calculate the value of x

Plug the equilibrium concentrations from the ICE table into the Kc expression: \[1.87 \times 10^{-6} = \frac{x(4.27 \times 10^{-4} + x)}{0.0111 - x}\] This is a quadratic equation in x. Solve the equation for x (mol of PH3) using the quadratic formula, substituting the given values.

Determine the equilibrium concentration of PH3

Once you have solved for x, you can use that to find the equilibrium concentration of PH3: \[[\mathrm{PH}_3] = \frac{x \text{ mol}}{1.500 \text{ L}}\] Calculate the value to obtain the equilibrium concentration of PH3.

Key concepts

Equilibrium Expression

In chemical equilibrium, the equilibrium expression is a mathematical representation of the balanced chemical equation. It relates the concentrations of the reactants and the products at equilibrium. For the given reaction:\[\text{PH}_3\text{BCl}_3(s) \rightleftharpoons \text{PH}_3(g) + \text{BCl}_3(g)\]Since \(\text{PH}_3\text{BCl}_3\) is a solid, it is not included in the equilibrium expression. Solids and liquids are typically not included because their concentrations do not change considerably. The expression for the equilibrium constant \(K_c\) is written in terms of the concentrations of the products and reactants raised to the power of their stoichiometric coefficients. Here, since both \(\text{PH}_3\) and \(\text{BCl}_3\) are gases, the equilibrium expression is:\[K_c = [\text{PH}_3][\text{BCl}_3]\]It is crucial to understand this expression as it allows us to calculate the extent to which a reaction will proceed to form products when it reaches equilibrium.

Gas Constant

The gas constant, denoted as \(R\), is a crucial constant that appears in the equation relating the equilibrium constants \(K_p\) and \(K_c\). The formula is:\[K_p = K_c(RT)^{\Delta n}\]where:- \(R\) is the ideal gas constant. Its value is \(0.0821 \text{L atm/mol K}\).- \(T\) is the temperature in Kelvin.- \(\Delta n\) represents the change in moles of gas during the reaction.The gas constant \(R\) allows for the conversion between the equilibrium constants \(K_p\), which is for gases in terms of partial pressures, and \(K_c\), which is in terms of molarity concentrations. It is essential in calculations involving gases, because it bridges the relationship between pressure and concentration.

ICE Table

An ICE table is an invaluable tool in equilibrium calculations, particularly useful for visualizing changes in concentrations. ICE stands for Initial, Change, and Equilibrium. To illustrate its use:1. **Initial:** This row indicates the starting quantities before any reaction occurs. For instance, initially, \(\text{PH}_3\) is absent in the mixture.2. **Change:** This reflects how the quantities change as the reaction moves toward equilibrium. Typically, it shows shifts in moles, for example, if \(x\) moles of \(\text{BCl}_3\) and \(\text{PH}_3\) form.3. **Equilibrium:** This final row provides the concentrations or pressures at equilibrium.For example, an ICE table for the problem would appear as follows:- \(\text{PH}_3\text{BCl}_3(s)\): Initial 0.0111 mol, Change \(-x\), Equilibrium \(0.0111 - x\)- \(\text{PH}_3(g)\): Initial 0 mol, Change \(+x\), Equilibrium \(x\)- \(\text{BCl}_3(g)\): Initial \(4.27 \times 10^{-4}\) mol, Change \(+x\), Equilibrium \(4.27 \times 10^{-4} + x\)Employing an ICE table ensures a structured method to determine equilibrium concentrations and solve for unknowns in equilibrium problems.

Equilibrium Concentration

The equilibrium concentration is the concentration of reactants and products in a chemical reaction that have reached a state of balance. It is the point where the forward and reverse reactions occur at the same rate, meaning there is no net change in the concentration of either reactants or products over time.To find the equilibrium concentration of \(\text{PH}_3\), we first need to solve for \(x\) using the ICE table, which represents the change in moles from initial to equilibrium. The quadratic formula often comes in handy here:- Use the expression \[K_c = \frac{[\text{PH}_3][\text{BCl}_3]}{[\text{PH}_3\text{BCl}_3]}\]- Substitute equilibrium values from the ICE table.- Solve the resulting quadratic equation.Once \(x\) is determined, calculate the \(\text{equilibrium concentration}\) of \(\text{PH}_3\) by dividing \(x\) by the volume of the reaction vessel, for this example, \(1.500 \, \text{L}\). Knowing how to calculate equilibrium concentrations enables you to predict concentrations at which reactions stop changing, which is very useful in many practical and theoretical chemistry applications.