/*! 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 37 At \(100^{\circ} \mathrm{C}\) th... [FREE SOLUTION] | 91影视

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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) \(\left[\mathrm{COCl}_{2}\right]=2.00 \times 10^{-3} \mathrm{M}, \quad[\mathrm{CO}]=3.3 \times 10^{-6} \mathrm{M}\) \(\left[\mathrm{Cl}_{2}\right]=6.62 \times 10^{-6} M ; \quad\) (b) \(\left[\mathrm{COCl}_{2}\right]=4.50 \times 10^{-2} M\) \([\mathrm{CO}]=1.1 \times 10^{-7} \mathrm{M},\left[\mathrm{Cl}_{2}\right]=2.25 \times 10^{-6} \mathrm{M} ;\) (c) \(\left[\mathrm{COCl}_{2}\right]=\) \(0.0100 M,[\mathrm{CO}]=\left[\mathrm{Cl}_{2}\right]=1.48 \times 10^{-6} \mathrm{M}\)

Short Answer

Expert verified
(a) The reaction will shift to the left, forming more reactants (COCl鈧). (b) The reaction will shift to the right, forming more products (CO and Cl鈧). (c) The reaction is already at equilibrium.

Step by step solution

01

Write down the balanced chemical equation and the equilibrium expression

We are given the balanced chemical equation: COCl鈧(g) 鈬 CO(g) + Cl鈧(g) The equilibrium expression for the reaction is given by: \(K_c = \frac{[CO][Cl_2]}{[COCl_2]}\) For each mixture, we will calculate the reaction quotient (Qc) and compare it to the equilibrium constant (Kc) to determine if the mixture is at equilibrium or not.
02

Calculate Qc for mixture (a)

Given concentrations for mixture (a): [COCl鈧俔 = 2.00 脳 10鈦宦 M [CO] = 3.3 脳 10鈦烩伓 M [Cl鈧俔 = 6.62 脳 10鈦烩伓 M Calculate Qc for mixture (a): \(Q_c = \frac{[CO][Cl_2]}{[COCl_2]} = \frac{(3.3 \times 10^{-6})(6.62 \times 10^{-6})}{(2.00 \times 10^{-3})} = 1.09 \times 10^{-8}\)
03

Compare Qc to Kc for mixture (a)

Given Kc = 2.19 脳 10鈦宦光伆, Since Qc (1.09 脳 10鈦烩伕) > Kc (2.19 脳 10鈦宦光伆), the reaction will shift to the left, forming more reactants (COCl鈧).
04

Calculate Qc for mixture (b)

Given concentrations for mixture (b): [COCl鈧俔 = 4.50 脳 10鈦宦 M [CO] = 1.1 脳 10鈦烩伔 M [Cl鈧俔 = 2.25 脳 10鈦烩伓 M Calculate Qc for mixture (b): \(Q_c = \frac{[CO][Cl_2]}{[COCl_2]} = \frac{(1.1 \times 10^{-7})(2.25 \times 10^{-6})}{(4.50 \times 10^{-2})} = 5.46 \times 10^{-12}\)
05

Compare Qc to Kc for mixture (b)

Given Kc = 2.19 脳 10鈦宦光伆, Since Qc (5.46 脳 10鈦宦孤) < Kc (2.19 脳 10鈦宦光伆), the reaction will shift to the right, forming more products (CO and Cl鈧).
06

Calculate Qc for mixture (c)

Given concentrations for mixture (c): [COCl鈧俔 = 0.0100 M [CO] = [Cl鈧俔 = 1.48 脳 10鈦烩伓 M Calculate Qc for mixture (c): \(Q_c = \frac{[CO][Cl_2]}{[COCl_2]} = \frac{(1.48 \times 10^{-6})(1.48 \times 10^{-6})}{(0.0100)} = 2.19 \times 10^{-10}\)
07

Compare Qc to Kc for mixture (c)

Given Kc = 2.19 脳 10鈦宦光伆, Since Qc (2.19 脳 10鈦宦光伆) = Kc (2.19 脳 10鈦宦光伆), the reaction is already at equilibrium for mixture (c). In conclusion, for mixture (a) the reaction will shift to the left, for mixture (b) it will shift to the right, and for mixture (c), it is already at equilibrium.

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Key Concepts

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

Equilibrium Constant (Kc)
The equilibrium constant, denoted as \(K_c\), is a key concept in understanding chemical equilibrium. It defines the ratio of the concentrations of products to reactants, raised to the power of their stoichiometric coefficients, at equilibrium. For a given reaction such as - \(\text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g)\), the equilibrium constant expression is written as:- \[K_c = \frac{[\text{CO}][\text{Cl}_2]}{[\text{COCl}_2]}\]This equation helps determine the position of equilibrium; whether it favors the reactants or the products.
  • A \(K_c\) value much greater than one indicates that, at equilibrium, products are favored.
  • A \(K_c\) value much less than one suggests that reactants are favored.
Understanding \(K_c\) is crucial as it provides insight into how a reaction will behave under certain conditions. It is important to note that \(K_c\) remains constant for a given reaction at a specific temperature.
Reaction Quotient (Qc)
The reaction quotient, \(Q_c\), is a useful tool for predicting the direction in which a reaction will proceed at any given point in time. It is calculated using the same formula as the equilibrium constant, \(K_c\). However, \(Q_c\) is computed with initial concentrations:- \[Q_c = \frac{[\text{CO}][\text{Cl}_2]}{[\text{COCl}_2]}\]By comparing \(Q_c\) to \(K_c\), one can determine how close a system is to equilibrium:
  • If \(Q_c = K_c\), the system is at equilibrium.
  • If \(Q_c < K_c\), the system will shift towards the products to reach equilibrium.
  • If \(Q_c > K_c\), the system will shift towards the reactants to reach equilibrium.
Thus, \(Q_c\) serves as a snapshot to help predict how a reaction will adjust its concentrations in order to find equilibrium, providing practical insights into reaction dynamics.
Le Chatelier's Principle
Le Chatelier's Principle is a key concept in chemical equilibrium that describes how a system at equilibrium reacts to disturbances. According to this principle, if an external change is applied to a system at equilibrium, the system will adjust itself in such a way as to counteract that change and achieve a new equilibrium state. There are several common disturbances that can affect a chemical reaction:
  • **Concentration Change:** Increasing the concentration of a reactant or product will cause the system to shift in the direction that consumes the added substance. Conversely, decreasing one will cause the system to shift to produce more of that substance.
  • **Temperature Change:** An increase in temperature for an endothermic reaction will shift the equilibrium towards the products, while for an exothermic reaction, it will shift towards the reactants.
  • **Pressure Change (for gaseous reactions):** Increasing pressure by decreasing volume will cause the equilibrium to shift towards the side with fewer moles of gas.
Understanding Le Chatelier's Principle helps predict how equilibrium systems will respond to changes, providing a predictive insight into the behavior of chemical reactions under various conditions.

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Most popular questions from this chapter

Write the expression for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(3 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g)\) (b) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) (c) \(\mathrm{Ni}(\mathrm{CO})_{4}(g) \rightleftharpoons \mathrm{Ni}(s)+4 \mathrm{CO}(g)\) (d) \(\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)\) (e) \(2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(a q) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{Zn}(s)\)

In Section \(11.5\) we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor, and the corresponding expression for \(K_{p}\). (b) By using data in Appendix \(\mathrm{B}\), give the value of \(K_{p}\) for this reaction at \(30{ }^{\circ} \mathrm{C}\). (c) What is the value of \(K_{p}\) for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

For the equilibrium $$ \mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{BrCl}(g) $$ at \(400 \mathrm{~K}, K_{c}=7.0\). If \(0.25 \mathrm{~mol}\) of \(\mathrm{Br}_{2}\) and \(0.25 \mathrm{~mol}\) of \(\mathrm{Cl}_{2}\) are introduced into a \(1.0\) -L container at \(400 \mathrm{~K}\), what will be the equilibrium concentrations of \(\mathrm{Br}_{2}, \mathrm{Cl}_{2}\), and \(\mathrm{BrCl} ?\)

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\). An equilibrium mix- ture in a 2.00-L vessel is found to contain \(0.0406 \mathrm{~mol}\) \(\mathrm{CH}_{3} \mathrm{OH}, 0.170 \mathrm{~mol} \mathrm{CO}\), and \(0.302 \mathrm{~mol} \mathrm{H}_{2}\) at \(500 \mathrm{~K}\). Cal- culate \(K_{c}\) at this temperature.

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?

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