EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
7th Edition
ISBN: 8220106637203
Author: Chang
Publisher: YUZU
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Chapter 15, Problem 15.45QP

(a)

Interpretation Introduction

Interpretation:

To predict the change of equilibrium constant (Kc) given set of endothermic and exothermic equilibrium reactions.

Concept Introduction:

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant.  If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Free energy: The change in the standard energy of the system that occurs during a reaction is therefore equal to the change in the enthalpy (ΔH) of the system minus the change in the respective product of the temperature at the times of entropy of the system, the following equation G=H-(TS) or (ΔG=ΔH-TΔS).

Le Chatelier's Principle (Kp): The closed system is an increase in pressure, the equilibrium will shift towards the sides of the reaction with some moles of gas.  The decrease in pressure the equilibrium will shift towards the side of the reaction with high moles of gas.

Exothermic and endothermic reaction: This type of chemical reaction release energy by light or heat.  These type of chemical reactions that is accompanied by the absorption of heat.

(b)

Interpretation Introduction

Interpretation:

To predict the change of equilibrium constant (Kc) given set of endothermic and exothermic equilibrium reactions.

Concept Introduction:

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant.  If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Free energy: The change in the standard energy of the system that occurs during a reaction is therefore equal to the change in the enthalpy (ΔH) of the system minus the change in the respective product of the temperature at the times of entropy of the system, the following equation G=H-(TS) or (ΔG=ΔH-TΔS).

Le Chatelier's Principle (Kp): The closed system is an increase in pressure, the equilibrium will shift towards the sides of the reaction with some moles of gas.  The decrease in pressure the equilibrium will shift towards the side of the reaction with high moles of gas.

Exothermic and endothermic reaction: This type of chemical reaction release energy by light or heat.  These type of chemical reactions that is accompanied by the absorption of heat.

(c)

Interpretation Introduction

Interpretation:

To predict the change of equilibrium constant (Kc) given set of endothermic and exothermic equilibrium reactions.

Concept Introduction:

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant.  If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Free energy: The change in the standard energy of the system that occurs during a reaction is therefore equal to the change in the enthalpy (ΔH) of the system minus the change in the respective product of the temperature at the times of entropy of the system, the following equation G=H-(TS) or (ΔG=ΔH-TΔS).

Le Chatelier's Principle (Kp): The closed system is an increase in pressure, the equilibrium will shift towards the sides of the reaction with some moles of gas.  The decrease in pressure the equilibrium will shift towards the side of the reaction with high moles of gas.

Exothermic and endothermic reaction: This type of chemical reaction release energy by light or heat.  These type of chemical reactions that is accompanied by the absorption of heat.

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Chapter 15 Solutions

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO

Ch. 15.4 - Practice Exercise At 430°C. the equilibrium...Ch. 15.4 - Prob. 2PECh. 15.4 - Prob. 1RCCh. 15.4 - Prob. 2RCCh. 15.4 - Prob. 3PECh. 15 - Prob. 15.1QPCh. 15 - 15.2 Explain the difference between physical...Ch. 15 - 15.3 Briefly describe the importance of...Ch. 15 - 15.4 Consider the equilibrium system 3A ⇌ B....Ch. 15 - 15.5 Define homogeneous equilibrium and...Ch. 15 - Prob. 15.6QPCh. 15 - 15.7 Write equilibrium constant expressions for...Ch. 15 - 15.8 Write the expressions for the equilibrium...Ch. 15 - 15.9 Write the equilibrium constant expressions...Ch. 15 - 15.10 Write the equation relating Kc and KP and...Ch. 15 - Prob. 15.11QPCh. 15 - Prob. 15.12QPCh. 15 - 15.13 The equilibrium constant (Kc) for the...Ch. 15 - Prob. 15.14QPCh. 15 - 15.15 What is the KP at 1273°C for the...Ch. 15 - 15.16 The equilibrium constant KP for the...Ch. 15 - Prob. 15.17QPCh. 15 - Prob. 15.18QPCh. 15 - Prob. 15.19QPCh. 15 - Prob. 15.20QPCh. 15 - Prob. 15.21QPCh. 15 - 15.22 Ammonium carbamate, NH4CO2NH2, decomposes...Ch. 15 - Prob. 15.23QPCh. 15 - 15.24 Pure phosgene gas (COCl2), 3.00 × 10−2...Ch. 15 - Prob. 15.25QPCh. 15 - 15.26 A 2.50-mol quantity of NOCl was initially...Ch. 15 - 15.27 Define reaction quotient. How does it...Ch. 15 - Prob. 15.28QPCh. 15 - Prob. 15.29QPCh. 15 - Prob. 15.30QPCh. 15 - Prob. 15.31QPCh. 15 - Prob. 15.32QPCh. 15 - Prob. 15.33QPCh. 15 - Prob. 15.34QPCh. 15 - Prob. 15.35QPCh. 15 - Prob. 15.36QPCh. 15 - Prob. 15.37QPCh. 15 - Prob. 15.38QPCh. 15 - Prob. 15.39QPCh. 15 - Prob. 15.40QPCh. 15 - Prob. 15.41QPCh. 15 - Prob. 15.42QPCh. 15 - Prob. 15.43QPCh. 15 - Prob. 15.44QPCh. 15 - Prob. 15.45QPCh. 15 - 15.46 What effect does an increase in pressure...Ch. 15 - Prob. 15.47QPCh. 15 - Prob. 15.48QPCh. 15 - 15.49 Consider the reaction Comment on the...Ch. 15 - Prob. 15.50QPCh. 15 - Prob. 15.51QPCh. 15 - Prob. 15.53QPCh. 15 - Prob. 15.54QPCh. 15 - Prob. 15.55QPCh. 15 - Prob. 15.56QPCh. 15 - Prob. 15.57QPCh. 15 - 15.58 Baking soda (sodium bicarbonate) undergoes...Ch. 15 - 15.59 Consider the following reaction at...Ch. 15 - Prob. 15.60QPCh. 15 - Prob. 15.61QPCh. 15 - Prob. 15.62QPCh. 15 - Prob. 15.64QPCh. 15 - Prob. 15.65QPCh. 15 - Prob. 15.66QPCh. 15 - Prob. 15.67QPCh. 15 - Prob. 15.68QPCh. 15 - Prob. 15.69QPCh. 15 - Prob. 15.70QPCh. 15 - Prob. 15.71QPCh. 15 - Prob. 15.72QPCh. 15 - Prob. 15.73QPCh. 15 - Prob. 15.74QPCh. 15 - Prob. 15.75QPCh. 15 - Prob. 15.76QPCh. 15 - Prob. 15.78QPCh. 15 - Prob. 15.79QPCh. 15 - Prob. 15.81QPCh. 15 - Prob. 15.82QPCh. 15 - Prob. 15.83QPCh. 15 - Prob. 15.84QPCh. 15 - Prob. 15.85QPCh. 15 - Prob. 15.86QPCh. 15 - Prob. 15.89QPCh. 15 - Prob. 15.90QPCh. 15 - Prob. 15.91QPCh. 15 - Prob. 15.92QPCh. 15 - Prob. 15.93QPCh. 15 - 15.94 Consider the decomposition of ammonium...Ch. 15 - Prob. 15.95QPCh. 15 - 15.96 In 1899 the German chemist Ludwig Mond...Ch. 15 - Prob. 15.98QPCh. 15 - Prob. 15.99QPCh. 15 - Prob. 15.100QPCh. 15 - Prob. 15.101QPCh. 15 - Prob. 15.102QPCh. 15 - Prob. 15.103SPCh. 15 - Prob. 15.104SPCh. 15 - Prob. 15.105SPCh. 15 - Prob. 15.106SPCh. 15 - Prob. 15.107SPCh. 15 - Prob. 15.110SPCh. 15 - Prob. 15.111SPCh. 15 - Prob. 15.112SP
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