/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 At \(100^{\circ} \mathrm{C},\) t... [FREE SOLUTION] | 91Ó°ÊÓ

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At \(100^{\circ} \mathrm{C},\) the equilibrium constant for the reaction \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) has the value \(K_{c}=2.19 \times 10^{-10} .\) Are the following mixtures of \(\mathrm{COCl}_{2}, \mathrm{CO},\) and \(\mathrm{Cl}_{2}\) at \(100^{\circ} \mathrm{C}\) at equilibrium? If not, indicate the direction that the reaction must proceed to achieve equilibrium. (a) \(\left[\mathrm{COCl}_{2}\right]=[\mathrm{CO}]=1.00 \times 10^{-4} \mathrm{M},\left[\mathrm{Cl}_{2}\right]=7.2 \times 10^{-6} \mathrm{M}\) (b) \(\left[\mathrm{COCl}_{2}\right]=2.20 \times 10^{-2} \mathrm{M},[\mathrm{CO}]=2.2 \times 10^{-7} \mathrm{M}\) \(\left[\mathrm{Cl}_{2}\right]=3.0 \times 10^{-6} \mathrm{M}\) (c) \(\left[\mathrm{COCl}_{2}\right]=0.0100 \mathrm{M},[\mathrm{CO}]=\left[\mathrm{Cl}_{2}\right]=7.2 \times 10^{-6} \mathrm{M}\)

Short Answer

Expert verified
(a) Not at equilibrium, shifts left; (b) Not at equilibrium, shifts right; (c) Not at equilibrium, shifts left.

Step by step solution

01

Determine Reaction Quotient Formula

The reaction quotient \( Q_c \) for the reaction \( \mathrm{COCl}_2(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{Cl}_2(g) \) is given by the formula:\[Q_c = \frac{[\mathrm{CO}] \times [\mathrm{Cl}_2]}{[\mathrm{COCl}_2]}\] This formula helps to compare the current state of the system with the equilibrium state.
02

Calculate Qc for Mixture (a)

For mixture (a), substitute the given concentrations:\[Q_c = \frac{(1.00 \times 10^{-4})(7.2 \times 10^{-6})}{1.00 \times 10^{-4}}\]Simplifying gives \( Q_c = 7.2 \times 10^{-6} \).
03

Compare Qc with Kc for Mixture (a)

Compare \( Q_c = 7.2 \times 10^{-6} \) with \( K_c = 2.19 \times 10^{-10} \). Since \( Q_c > K_c \), the system is not at equilibrium and will shift to the left (towards the reactant) to reach equilibrium.
04

Calculate Qc for Mixture (b)

For mixture (b), substitute the given concentrations:\[Q_c = \frac{(2.2 \times 10^{-7})(3.0 \times 10^{-6})}{2.20 \times 10^{-2}}\]Simplifying gives \( Q_c \approx 3.0 \times 10^{-11} \).
05

Compare Qc with Kc for Mixture (b)

Compare \( Q_c = 3.0 \times 10^{-11} \) with \( K_c = 2.19 \times 10^{-10} \). Since \( Q_c < K_c \), the system is not at equilibrium and will shift to the right (towards the products) to reach equilibrium.
06

Calculate Qc for Mixture (c)

For mixture (c), substitute the given concentrations:\[Q_c = \frac{(7.2 \times 10^{-6})(7.2 \times 10^{-6})}{0.0100}\]Simplifying gives \( Q_c \approx 5.18 \times 10^{-9} \).
07

Compare Qc with Kc for Mixture (c)

Compare \( Q_c = 5.18 \times 10^{-9} \) with \( K_c = 2.19 \times 10^{-10} \). Since \( Q_c > K_c \), the system is not at equilibrium and will shift to the left (towards the reactant) to reach equilibrium.

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Key Concepts

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

Reaction Quotient
In chemical equilibrium, the reaction quotient, represented as \( Q_c \), is a helpful tool to determine whether a system is at equilibrium. The reaction quotient is calculated in a similar manner to the equilibrium constant \( K_c \), but uses the current concentrations of reactants and products in the formula. For our example reaction \( \text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g) \), it is calculated as follows:
\[Q_c = \frac{[\text{CO}] \times [\text{Cl}_2]}{[\text{COCl}_2]}\]
It's important to note that \( Q_c \) can change as a reaction progresses, unlike \( K_c \) which is constant at a given temperature.
After calculating \( Q_c \), it is compared to \( K_c \), the equilibrium constant, to assess the system's state:
  • If \( Q_c = K_c \), the system is at equilibrium.
  • If \( Q_c > K_c \), there are more products compared to reactants than at equilibrium, so the reaction will shift towards the reactants to reach equilibrium.
  • If \( Q_c < K_c \), the products are less than the equilibrium concentration, necessitating a shift towards product formation to attain equilibrium as in Mixture (b) from the example.
This comparison provides insights into the direction in which a reaction needs to adjust to achieve equilibrium.
Equilibrium Constant
The equilibrium constant \( K_c \) is a crucial concept in understanding chemical equilibrium. It represents the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients, at equilibrium. It's a fixed value at a given temperature.
For the reaction \( \text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g) \), the equilibrium constant is expressed as:
\[K_c = \frac{[\text{CO}] \times [\text{Cl}_2]}{[\text{COCl}_2]}\]
The value of \( K_c \) is given for the reaction at \( 100^{\circ} \text{C} \) as \( 2.19 \times 10^{-10} \). This small value indicates that at equilibrium, the concentrations of reactants are favored over products, as there are fewer molecules of \( \text{CO} \) and \( \text{Cl}_2 \) relative to \( \text{COCl}_2 \).
On comparing the calculated \( Q_c \) values for the mixtures in the exercise with \( K_c \), one can determine in which direction the reaction needs to shift. This helps in predicting how the system reacts to changes in concentration, temperature, and pressure, as it strives for equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle is a guiding concept for predicting how a chemical system at equilibrium will respond to changes. It states that if a change or stress is applied to a system at equilibrium, the system will adjust itself to counteract that change and restore a new equilibrium.
In the context of the reaction \( \text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g) \), if the concentration of one of the components is changed, the system will strive to adjust the concentrations to offset that change, thereby moving towards a new equilibrium state.
  • Addition of more \( \text{COCl}_2 \) will shift the reaction to the right (forming more products) to alleviate the stress.
  • Conversely, adding more \( \text{CO} \) or \( \text{Cl}_2 \) will push the reaction to the left (forming more reactant) to decrease the product concentration.
This principle also applies to changes in temperature and pressure, affecting the direction in which a reaction proceeds. It is a valuable tool for understanding and predicting the behavior of equilibria in chemical systems.

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Most popular questions from this chapter

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?

When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products? (a) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) at \(900 \mathrm{~K} K_{c}=2.24\) (b) \(\left.2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\right)\) at \(1300 \mathrm{~K} K_{p}=5.27 \times 10^{-5}\)

At \(800 \mathrm{~K},\) the equilibrium constant for the reaction \(\mathrm{A}_{2}(g) \rightleftharpoons 2 \mathrm{~A}(g)\) is \(K_{c}=3.1 \times 10^{-4}\) (a) Assuming both forward and reverse reactions are elementary reactions, which rate constant do you expect to be larger, \(k_{f}\) or \(k_{r} ?\) (b) If the value of \(k_{f}=0.27 \mathrm{~s}^{-1}\), what is the value of \(k_{r}\) at \(800 \mathrm{~K}\) ? (c) Based on the nature of the reaction, do you expect the forward reaction to be endothermic or exothermic? (d) If the temperature is raised to \(1000 \mathrm{~K},\) will the reverse rate constant \(k_{r}\) increase or decrease? Will the change in \(k_{r}\) be larger or smaller than the change in \(k_{f}\) ?

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 5.00-L flask at \(310 \mathrm{~K}\), \(40 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{P}\) for this reaction at \(310 \mathrm{~K}\). (c) According to Le Châtelier's principle, would the percent of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes increase, decrease or stay the same if the mixture was transferred to a 1.00-L vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when 2.00 mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 1.00-L vessel at \(310 \mathrm{~K}\).

Bromine and hydrogen react in the gas phase to form hydrogen bromide: \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\). The reaction enthalpy is \(\Delta H^{\circ}=-6 \mathrm{~kJ} .(\mathbf{a})\) To increase the equilibrium yield of hydrogen bromide would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen bromide by controlling the pressure of this reaction? If so, would high or low pressure favor formation of \(\mathrm{HBr}(g)\) ?

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