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For the equilibrium $$ \mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g) $$ \(K_{p}=0.052\) at \(60^{\circ} \mathrm{C}\). (a) Calculate \(K_{c}\). (b) Some solid \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) is added to a closed \(0.500\) -L vessel at \(60^{\circ} \mathrm{C}\); the vessel is then charged with \(0.0128\) mol of \(\mathrm{BCl}_{3}(g) .\) What is the equilibrium concentration of \(\mathrm{PH}_{3}\) ?

Short Answer

Expert verified
The equilibrium constant Kc for the given reaction is approximately \(2.47\times10^{-7}\) and the equilibrium concentration of PH3 is approximately \(9.65\times10^{-6} \mathrm{M}\).

Step by step solution

01

Write the relationship between Kp and Kc

To calculate Kc given Kp, we use the relationship: \[ K_{p} = K_{c} (RT)^{\Delta{n}} \] where R is the ideal gas constant, T is the temperature in Kelvin, and Δn is the change in moles of gas for the reaction.
02

Calculate Δn

To find the change in moles of gas (Δn), we need to look at the balanced chemical equation: \[ PH_3BCl_3(s) \rightleftharpoons PH_3(g) + BCl_3(g) \] For each mole of solid reactant, 1 mole of PH3 gas and 1 mole of BCl3 gas are produced. Hence, Δn = (1 + 1) - 0 = 2
03

Convert temperature to Kelvin

The temperature given is in degree Celsius, we need to convert it into Kelvin before using it in the equation. Temperature in Kelvin = 60 + 273.15 = 333.15 K
04

Calculate Kc using Δn, temperature and Kp

Now, we can plug in the values into the relationship we established in step 1: \[ K_{p} = K_{c} (RT)^{\Delta{n}} \] Substitute the values : \(0.052 = K_{c} \times (0.08206 \times 333.15)^{2} \) Solve for Kc: \[ K_{c} = \frac{0.052}{(0.08206 \times 333.15)^{2}} \] \[ K_{c} \approx 2.47\times10^{-7} \] #Part B: Calculate the PH3 equilibrium concentration#
05

Write the Kc expression for the reaction

Knowing Kc, we can establish an equilibrium expression for the reaction: \[ K_c = \frac{[PH_3][BCl_3]}{[PH_3BCl_3]}\] However, since PH3BCl3 is a solid, its concentration does not change during the reaction and is not included in the Kc expression. Thus, the expression becomes: \[ K_c = [PH_3][BCl_3] \]
06

Set up an ICE table

Now, we will set up an Initial, Change, and Equilibrium (ICE) table to find the equilibrium concentrations of PH3 and BCl3: | Species | Initial | Change | Equilibrium | | ------------ | ------- | -------- | ----------- | | PH3(g) | 0 | +x | x | | BCl3(g) | 0.0256 | -x | 0.0256 - x | Initially, we have no PH3 present. We're told the vessel is charged with 0.0128 mol of BCl3, and the volume of the vessel is 0.500 L, so the initial concentration of BCl3 is 0.0128 mol/0.500 L = 0.0256 M.
07

Substitute the equilibrium concentrations into the Kc expression

Plug the equilibrium concentrations into the Kc expression: \[ K_c = [PH_3][BCl_3] \] \[ 2.47\times10^{-7} = x(0.0256-x) \]
08

Solve for x, the equilibrium concentration of PH3

Since Kc is very small, we can assume that x is negligible compared to 0.0256: \[ 2.47\times10^{-7} \approx x(0.0256) \] \[ x \approx \frac{2.47\times10^{-7}}{0.0256} \] \[ x \approx 9.65\times10^{-6} \mathrm{M} \] The equilibrium concentration of PH3 is approximately \(9.65\times10^{-6} \mathrm{˜M}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant (Kc)
Understanding the equilibrium constant, denoted as Kc, is central to mastering chemical equilibrium. Kc is a numerical value that represents the ratio of the concentration of the products raised to the power of their coefficients to the concentration of the reactants raised to the power of their coefficients at equilibrium for a reversible reaction in a closed system.

In the mathematical term, for a general reaction: \[ aA + bB \rightleftharpoons cC + dD \]
The equilibrium constant in terms of concentration (Kc) is expressed as:\[ K_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a}[B]^{b}} \]
where [A], [B], [C], and [D] represent the molar concentrations of the reactants and products. It's important to note that only gases and aqueous solutions are included in this expression - solids and pure liquids, being invariable in concentration, are omitted.

Furthermore, Kc is highly temperature-dependent and remains constant only at a given temperature. If we know Kc and the initial concentrations of reactants and products, we can predict the direction in which the reaction will proceed to reach equilibrium.
Equilibrium Constant (Kp)
For reactions involving gases, we often use the equilibrium constant in terms of partial pressure, represented by Kp. It's directly related to Kc but uses the pressures of the gaseous reactants and products instead of their concentrations.

The Kp for a reaction is given by the expression:

\[ K_{p} = \frac{P_{C}^{c} P_{D}^{d}}{P_{A}^{a}P_{B}^{b}} \]
where P represents the partial pressure of the gases involved. The relationship between Kp and Kc is delineated by the formula:

\[ K_{p} = K_{c}(RT)^{\Delta n} \]
where R is the universal gas constant, T is the temperature in Kelvin, and \( \Delta n \) is the difference in the number of moles of gaseous products and reactants. The exercise we've looked at illustrates how we can calculate Kc from Kp using this equation, acknowledging the crucial role temperature and mole changes play in this context.
ICE Table
Getting to grips with an ICE (Initial, Change, Equilibrium) table is essential for students tackling equilibrium problems. This straightforward yet powerful tool helps systematically determine the equilibrium concentrations or pressures for all species in a reaction.

An ICE table lays out the initial concentrations or pressures, the changes that occur as reactants proceed to products, and the equilibrium concentrations or pressures. To construct an ICE table, you begin by filling in initial concentrations or partial pressures, then express the changes that occur in terms of an unknown variable 'x', and finally, define the equilibrium concentrations or pressures in terms of 'x'.

Once you've set up your ICE table, you can use the known equilibrium constant to solve for 'x', thus finding the equilibrium concentrations. This method is clearly demonstrated in the exercise, where the ICE table is crucial for calculating the equilibrium concentration of PH3(g).
Le Châtelier's Principle
Le Châtelier's Principle is a vital concept in chemical equilibrium that enables us to predict how a system at equilibrium responds to external changes. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change and re-establish equilibrium.

Several factors can affect the equilibrium position, including changes in concentration, pressure, volume, and temperature. For example, increasing the concentration of reactants will shift the equilibrium towards the products to lower the concentration of reactants, while an increase in temperature for an exothermic reaction will shift the equilibrium towards the reactants as the system seeks to absorb the extra heat.

In practice, Le Châtelier's Principle explains why adding more solid PH3BCl3 does not affect the equilibrium position in our exercise: solids do not appear in the equilibrium constant expression, and their concentration is considered unchanged. This principle ensures predictability in chemical processes, highlighting its significance in both academic and industrial applications.

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Most popular questions from this chapter

At \(700 \mathrm{~K}\) the equilibrium constant for the reaction $$ \mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(\mathrm{s})+2 \mathrm{Cl}_{2}(g) $$ is \(K_{p}=0.76\). A flask is charged with \(2.00 \mathrm{~atm}\) of \(\mathrm{CCl}_{4}\), which then reaches equilibrium at \(700 \mathrm{~K}\). (a) What fraction of the \(\mathrm{CCl}_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?\) (b) What are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by increasing the temperature and increasing the volume of the reaction vessel. (a) What can you conclude about the reaction from the influence of temperature on the equilibrium? (b) What can you conclude from the influence of increasing the volume?

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.16), \(K_{p}\) for the reaction $$ 2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) $$ is about \(1 \times 10^{-13}\). Assuming that the exhaust gas (total pressure \(1 \mathrm{~atm}\) ) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2}\), and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the above reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the reaction above?

For the equilibrium $$ 2 \operatorname{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) $$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C}\). If \(0.025 \mathrm{~atm}\) of \(\mathrm{IBr}\) is placed in a \(2.0\) - \(\mathrm{L}\) container, what is the partial pressure of this substance after equilibrium is reached?

A flask is charged with \(1.500\) atm of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(1.00 \mathrm{~atm} \mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C}\), and the following equilibrium is achieved: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is \(0.512\) atm. (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (b) Calculate the value of \(K_{p}\) for the reaction.

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