/*! 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 83 Nitric oxide (NO) reacts readily... [FREE SOLUTION] | 91Ó°ÊÓ

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Nitric oxide (NO) reacts readily with chlorine gas as follows: $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g) $$ At \(700 \mathrm{~K}\), the equilibrium constant \(K_{p}\) for this reaction is \(2.6 \times 10^{-3}\). Predict the behavior of each of the following mixtures at this temperature and indicate whether or not the mixtures are at equilibrium. If not, state whether the mixture will need to produce more products or reactants to reach equilibrium. (a) \(P_{\mathrm{NO}}=20.3 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=20.3 \mathrm{kPa}, P_{\mathrm{NOCl}}=20.3 \mathrm{kPa}\) (b) \(P_{\mathrm{NO}}=25.33 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=15.2 \mathrm{kPa}, P_{\mathrm{NOCl}}=2.03 \mathrm{kPa}\) (c) \(P_{\mathrm{NO}}=15.2 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=42.6 \mathrm{kPa}, P_{\mathrm{NOCl}}=5.07 \mathrm{kPa}\)

Short Answer

Expert verified
(a) Mixture shifts to reactants. (b) Mixture shifts to products. (c) Mixture is at equilibrium.

Step by step solution

01

Identify Q for Case (a)

For the first mixture, calculate the reaction quotient \( Q_p \). The equation for \( Q_p \) based on partial pressures is:\[ Q_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 \cdot P_{\text{Cl}_2}} \]Substitute the given pressures:\[ Q_p = \frac{(20.3)^2}{(20.3)^2 \cdot 20.3} = \frac{412.09}{8328.03} \approx 0.0495 \]
02

Compare Q and K for Case (a)

Compare the calculated \( Q_p = 0.0495 \) with the equilibrium constant \( K_p = 2.6 \times 10^{-3} \). Since \( Q_p > K_p \), the reaction mixture will shift to favor the reactants to reach equilibrium.
03

Identify Q for Case (b)

Calculate \( Q_p \) for the second mixture:\[ Q_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 \cdot P_{\text{Cl}_2}} \]Substitute the given pressures:\[ Q_p = \frac{(2.03)^2}{(25.33)^2 \cdot 15.2} = \frac{4.1209}{9751.02} \approx 4.23 \times 10^{-4} \]
04

Compare Q and K for Case (b)

Here, \( Q_p = 4.23 \times 10^{-4} \) and \( K_p = 2.6 \times 10^{-3} \). Since \( Q_p < K_p \), the reaction will produce more products to reach equilibrium.
05

Identify Q for Case (c)

Calculate \( Q_p \) for the third mixture:\[ Q_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 \cdot P_{\text{Cl}_2}} \]Substitute the given pressures:\[ Q_p = \frac{(5.07)^2}{(15.2)^2 \cdot 42.6} = \frac{25.7049}{9853.824} \approx 0.00261 \]
06

Compare Q and K for Case (c)

For this case, \( Q_p = 0.00261 \) and \( K_p = 2.6 \times 10^{-3} \). Given that \( Q_p \approx K_p \), this mixture is at equilibrium.

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Key Concepts

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

Reaction Quotient - Understanding Q
In chemical reactions, the reaction quotient, denoted as \( Q_p \), is a crucial concept to understand. It represents the ratio of the concentrations (or partial pressures) of products to reactants at any given point in time during a reaction. This is important because it allows us to determine whether a system is at equilibrium, or if it still needs adjustments.
For the given reaction of nitric oxide (NO) with chlorine gas (\(Cl_2\)), the expression for the reaction quotient \( Q_p \) is:
  • \[ Q_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 \cdot P_{\text{Cl}_2}} \]
When the reaction starts, we can calculate \(Q_p\) using the initial partial pressures of each gas. This value helps identify the state of the system. If \( Q_p \) is significantly different from the equilibrium constant \( K_p \), the system will adjust itself to reach equilibrium by shifting towards either the reactants or the products. If \( Q_p \) equals \( K_p \), the system is already at equilibrium.
This concept helps in predicting the direction of the reaction, guiding chemists on how a reaction needs to be shifted to reach equilibrium conditions.
Equilibrium Constant - Keeping the Balance
The equilibrium constant, \( K_p \), is a fundamental constant specific to a particular reaction at a given temperature. It reflects the ratio of the concentrations or partial pressures of products to reactants when the system is at equilibrium.
For the reaction between nitric oxide and chlorine gas, \( K_p \) is given as \(2.6 \times 10^{-3}\) at 700 K. This constant illustrates the point at which the rates of the forward and backward reactions are balanced, meaning no net change occurs in the concentration of reactants and products.
  • If \( Q_p < K_p \), the reaction will shift towards the products. This means more products will be formed until equilibrium is achieved.
  • If \( Q_p > K_p \), the system will favor the formation of reactants, reducing the product concentration to reach equilibrium.
  • If \( Q_p = K_p \), the reaction is at equilibrium.
Understanding \(K_p\) and how it works in relation to the reaction quotient is key to predicting how a chemical system will behave.
Le Chatelier's Principle - Shift Happens
Le Chatelier's Principle is an essential concept in understanding how systems at equilibrium respond to changes in conditions. This principle states that if a system at equilibrium is subjected to a change, such as a change in concentration, temperature, or pressure, the system will adjust in a way that counteracts the imposed change, trying to restore a new equilibrium.
In the reaction involving \( NO \) and \( Cl_2 \), if the partial pressures of reactants or products are altered, these changes can cause the reaction to shift in order to re-achieve equilibrium:
  • If more reactant or product is added, the reaction shifts to use up the added substance.
  • If pressure is increased by reducing volume, the system shifts towards the side with fewer gas molecules.
  • Temperature changes can shift equilibrium depending on the exothermic or endothermic nature of the reaction.
Applying Le Chatelier’s Principle helps predict how the reaction conditions need to be changed to maintain balance or to produce more of a desired product.

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Most popular questions from this chapter

For the equilibrium $$ 2 \operatorname{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\operatorname{Br}_{2}(g) $$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C}\). If \(5.07 \mathrm{kPa}\) of IBr is placed in a 10.0-L container, what is the partial pressure of all substances after equilibrium is reached?

If \(K_{c}=1\) for the equilibrium \(3 \mathrm{~A}(g) \rightleftharpoons 2 \mathrm{~B}(g),\) what is the relationship between [A] and [B] at equilibrium?

Calculate \(K_{c}\) at \(900 \mathrm{~K}\) for \(2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\) if \(K_{p}=0.0572\) at this temperature.

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 \(k_{f}\) ?

The reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) has \(K_{p}=\) \(8.59 \times 10^{-4}\) at \(300^{\circ} \mathrm{C}\). A flask is charged with \(50.7 \mathrm{kPa}\) \(\mathrm{PCl}_{3}, 50.7 \mathrm{kPa} \mathrm{Cl}_{2}\), and \(20.3 \mathrm{kPa} \mathrm{PCl}_{5}\) at this temperature. (a) Use the reaction quotient to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium partial pressures of the gases. \((\mathbf{c})\) What effect will increasing the volume of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing the temperature of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

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