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Four moles of \(\mathrm{PCl}_{5}\) are placed in a 2 -liter flask. When the following equilibrium is established $$ \mathrm{PCI}_{5} \rightleftarrows \mathrm{PCI}_{3}+\mathrm{CI}_{2} $$ the flask is found to contain \(0.8\) mole of \(\mathrm{Cl}_{2}\). What is the equilibrium constant?

Short Answer

Expert verified
The equilibrium constant (K) for the reaction \(\mathrm{PCl}_{5} \rightleftarrows \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\) is 0.533.

Step by step solution

01

Write the equilibrium expression

For the given reaction, the equilibrium constant expression (K) can be written as: $$ K = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_5]} $$
02

Determine initial concentrations

The initial moles of PCl5 are given as 4 moles and it is placed in a 2-liter flask. So, the initial concentration of PCl5 is: $$ [\mathrm{PCl}_5]_{initial} = \frac{4\,\text{moles}}{2\,\text{liters}} = 2\,\text{M} $$ Initially, PCl3 and Cl2 are not present, so their concentrations are zero.
03

Calculate concentration changes

Since the reaction is at equilibrium and 0.8 moles of Cl2 are present in the flask, the change in concentration for each specie can be represented as follows: Change in PCl5: \(-0.8\,\text{M}\) Change in PCl3: \(+0.8\,\text{M}\) Change in Cl2: \(+0.8\,\text{M}\)
04

Determine equilibrium concentrations

Using the changes in concentrations calculated in Step 3, we can determine the concentrations at equilibrium. $$ [\mathrm{PCl}_5]_{equilibrium} = 2\,\text{M} - 0.8\,\text{M} = 1.2\,\text{M}\\ [\mathrm{PCl}_3]_{equilibrium} = 0 + 0.8\,\text{M} = 0.8\,\text{M}\\ [\mathrm{Cl}_{2}]_{equilibrium} = 0 + 0.8\,\text{M} = 0.8\,\text{M} $$
05

Calculate the equilibrium constant K

Now that we have the equilibrium concentrations, we can plug them into the equilibrium expression to calculate K: $$ K = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_5]} = \frac{(0.8)(0.8)}{(1.2)} = \frac{0.64}{1.2} = 0.533 $$ The equilibrium constant for the given reaction is 0.533.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is fundamental to grasping the behavior of reactions that can occur in both the forward and reverse directions. When a reaction reaches the state of chemical equilibrium, the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time.

Imagine a busy marketplace where people are constantly buying and selling goods. At some point, the amount of goods being bought is equal to the amount being sold, balancing the transactions. Similarly, in a chemical reaction at equilibrium, the rate at which products are formed from the reactants equals the rate at which products revert to reactants.

Key Points:
  • At equilibrium, concentrations of reactants and products remain constant.
  • It implies a dynamic balance, not a static one—molecules are always reacting, just at equal rates for both the forward and reverse reactions.
  • Changes to conditions such as temperature, pressure, or concentration can disrupt this balance, potentially leading to a new equilibrium state (as explained by Le Chatelier's principle).
Equilibrium Expression Derivation
To calculate the equilibrium constant (often denoted as K or Keq), it is essential to derive the correct equilibrium expression based on the balanced chemical equation. The equilibrium constant is a numerical representation of the ratio of the concentration of products raised to their coefficients, to the concentration of reactants also raised to their coefficients.

In simplified terms, the equilibrium expression for a generic chemical reaction is given by:
$$K = \frac{[\text{Products}]}{[\text{Reactants}]}$$
For a reaction such as \(aA + bB \rightleftarrows cC + dD\), the equilibrium constant expression would be:
$$K = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$$
It's crucial not to include solids or liquids in the equilibrium expression since their concentrations are considered constant and do not affect the equilibrium state.

Key Points:
  • The equilibrium expression is derived from the balanced chemical equation of the reaction.
  • Only species in the gaseous or aqueous state are included in the expression as their concentrations can change.
  • Coefficients in the balanced equation become exponents in the equilibrium expression.
Concentration Calculation
Calculating the concentrations of reactants and products at equilibrium is a vital step in determining the equilibrium constant. Initial concentrations are typically provided or can be calculated from given amounts and volumes. As the system reaches equilibrium, these concentrations change due to the consumption and formation of species according to the stoichiometry of the balanced equation.

For example, if one mole of a reactant leads to the formation of one mole of a product, then a decrease in the reactant’s concentration by a certain amount will result in an equal increase in the product’s concentration. This is based on the assumption of a one-to-one mole ratio, which may differ according to the chemical equation.

Key Points:
  • To begin, calculate initial concentrations, usually by dividing the number of moles by the volume of the solution.
  • Track changes in concentration by considering the stoichiometry of the reaction.
  • Determine the equilibrium concentrations by accounting for these changes.
  • Concentrations of pure solids and liquids are not part of calculations as their concentrations do not change.
Le Chatelier's Principle
Le Chatelier's principle provides a predictive tool for understanding how changes in conditions, such as concentration, temperature, and pressure, affect a chemical system at equilibrium. According to this principle, if an external change is applied to a system at equilibrium, the system will adjust itself to counteract that change and establish a new equilibrium.

For instance, adding more reactants to the system will result in the formation of more products until a new equilibrium is attained. Conversely, increasing the temperature of an exothermic reaction will favor the formation of reactants because the system tries to absorb the added heat.

Key Points:
  • Le Chatelier's principle applies to systems at chemical equilibrium.
  • The principle predicts the shift in equilibrium in response to external changes.
  • A new equilibrium is established after the system adjusts to minimize the effect of the disturbance.
  • This understanding helps chemists control the yields of reactions and design processes more efficiently.

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Most popular questions from this chapter

At \(986^{\circ} \mathrm{C}, \mathrm{K}=1.60\) for the reaction, \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftarrows\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) .\) If you inject one mole each of \(\mathrm{H}_{2}, \mathrm{CO}_{2}\) \(\mathrm{H}_{2} \mathrm{O}\), and CO simultaneously in a 20 -liter box at time \(\mathrm{t}=0\) and allow them to equilibrate at \(986^{\circ} \mathrm{C}\), what will be the final concentrations of all the species? What would happen to these concentrations if additional \(\mathrm{H}_{2}\) was injected and a new equilibrium was established?

One of the two most important uses of ammonia is as a reagent in the first step of the Osfwald process, a synthetic route for the production of nitric acid. This first step proceeds according to the equation $$ 4 \mathrm{NH}_{3}(\mathrm{~g})+50_{2}(\mathrm{~g}) \rightleftarrows 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ What is the expression for the equilibrium constant of this reaction?

At \(395 \mathrm{~K}\), chlorine and carbon monoxide react to form phosgene, \(\mathrm{COCI}_{2}\), according to the equation $$ \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftarrows \mathrm{COCI}_{2}(\mathrm{~g}) $$ The equilibrium partial pressures are \(\mathrm{PCL}_{2}=0.128 \mathrm{~atm}\), \(\mathrm{P}_{\mathrm{CO}}=0.116 \mathrm{~atm}\), and \(\mathrm{P}_{(\mathrm{COCl}) 2}=0.334 \mathrm{~atm} .\) Determine the equilibrium constant \(\mathrm{K}_{\mathrm{p}}\) for the dissociation of phosgene and the degree of dissociation at \(395 \mathrm{~K}\) under a pressure of \(1 \mathrm{~atm} .\)

For the reaction $$ 2 \mathrm{HI}(\mathrm{g}) \rightleftarrows \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) $$ the value of the equilibrium constant at \(700 \mathrm{~K}\) is \(0.0183 .\) If \(3.0\) moles of \(\mathrm{HI}\) are placed in a 5 -liter vessel and allowed to decompose according to the above equation, what percentage of the original HI would remain undissociated at equilibrium?

At \(986^{\circ} \mathrm{C}\), you have the following equilibrium: \(\mathrm{C} \mathrm{O}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightleftarrows \mathrm{C} 0(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) Initially, \(49.3\) mole percent \(\mathrm{CO}_{2}\) is mixed with \(50.7 \mathrm{~mole}\) per cent \(\mathrm{H}_{2}\). At equilibrium, you find \(21.4\) mole percent \(\mathrm{CO}_{2}\), \(22.8\) mole percent \(\mathrm{H}_{2}\), and \(27.9 \mathrm{~mole}\) percent of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\). Find \(\mathrm{K}\). If you start with a mole percent ratio of \(60: 40\) \(\mathrm{CO}_{2}\) to \(\mathrm{H}_{2}\), find the equilibrium concentrations of both reactants and products.

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