General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 18, Problem 18.51QP

(a)

Interpretation Introduction

Interpretation:

To derive the entropy, enthalpy (ΔHοandΔSο) values for given gas phase equilibrium reaction (K1 and K2) at 25°C

Concept Information:

Gibbs free energy (G): The thermodynamic quantity to the (ΔG0) enthalpy of a system process and minutes the product of entropy and the absolute temperature.  The energy associated with a chemical reaction that can be used to do work.  The free energy of a system is the sum of its enthalpy (H) plus the product of the temperature (Kelvin) and the entropy (S) of the system.

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(ΔG0): The entropy is second law of thermodynamics it indicated for (ΔG0) symbol, the many of chemical reactions cause changes in entropy and it plays on important role in determining in which direction (forward and backward) a chemical reaction spontaneously proceeds.

(a)

Expert Solution
Check Mark

Explanation of Solution

Derive the equilibrium constants and different temperature for give standard state conditions (ΔHοandΔSο) equation.

Let us consider the following thermodynamic equations,

  N2O4(g)2NO2(g)ΔHο=58.0KJ/molGiven the two differnt temprature (T1andT2)ΔG1ο=ΔHοT1ΔSο=RTInK1----[1]ΔG2ο=ΔHο-T2ΔSο=-RTInK2-[2]RearrangingEquations(1)and(2)wegetln K1=-ΔHοRT1+ΔSοR[3]ln K2=-ΔHοRT2+ΔSοR[4]Subtractingequation(3)and(4)givesln K2ln K1=(-ΔHοRT2+ΔSοR)(-ΔHοRT1+ΔSοR)lnK2K1=ΔHοR(1T11T2)lnK2K1=ΔHοR(T2-T1T1T2)

(b)

Interpretation Introduction

Interpretation:

To derive the equilibrium constant (K) values for given gas phase equilibrium reaction (K1 and K2) at 25°C, 65°C

Concept Information:

Thermodynamics is the branch of science that relates heat and energy in a system.  The laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy. Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

ΔG = ΔΗ- TΔS

Where,

  ΔG  is the change in free energy of the system

  ΔΗ is the change in enthalpy of the system

  T is the absolute value of the temperature

  ΔS is the change in entropy in the system.

The equilibrium constant can be calculated by using following formula,

  lnK2K1=ΔHοR(T2-T1T1T2)

(b)

Expert Solution
Check Mark

Explanation of Solution

Using the above equation that we can derived, we can calculate the equilibrium constant at 65°C

  lnK2K1=ΔHοR(T2-T1T1T2)

  Hence calculated the equilibrium constant at650CN2O4(g)2NO2(g)ΔHο=58.0KJ/molGiven the statement of values are K1=4.63×10-3,T1=273+25=298K,K2=?T2=273+65=338KHence we derivedthe equilibrum constantvalues of(K2)lnK24.63×10-3=58.0×1038.314J/K×mol(338K-298K(338K)(298K))lnK24.63×10-3=58.0×1038.314J/K×mol(40100724)lnK24.63×10-3=58.0×1038.314J/K×mol(3.9717×104)lnK24.63×10-3=(6.9761×105)(3.9717×104)lnK24.63×10-3=2.77K2=0.074

K2>K1

As we would predict for a positive ΔH°, if  increase in temperature will shift the equilibrium toward the endothermic reaction, that is decomposition of N2O4  

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Chapter 18 Solutions

General Chemistry

Ch. 18.6 - Practice Exercise Calculate the equilibrium...Ch. 18.6 - Prob. 2PECh. 18.6 - Prob. 3PECh. 18.6 - Prob. 1RCCh. 18 - Prob. 18.1QPCh. 18 - Prob. 18.2QPCh. 18 - Prob. 18.3QPCh. 18 - Prob. 18.4QPCh. 18 - Prob. 18.5QPCh. 18 - Prob. 18.7QPCh. 18 - Prob. 18.8QPCh. 18 - Prob. 18.9QPCh. 18 - 18.10 Arrange the following substances (1 mole...Ch. 18 - Prob. 18.11QPCh. 18 - Prob. 18.12QPCh. 18 - Prob. 18.13QPCh. 18 - 18.14 State whether the sign of the entropy...Ch. 18 - 18.15 Define free energy. What are its units? Ch. 18 - 18.16 Why is it more convenient to predict the...Ch. 18 - 18.17 Calculate ΔG° for the following reactions at...Ch. 18 - 18.18 Calculate ΔG° for the following reactions at...Ch. 18 - Prob. 18.19QPCh. 18 - Prob. 18.20QPCh. 18 - Prob. 18.21QPCh. 18 - Prob. 18.22QPCh. 18 - Prob. 18.23QPCh. 18 - 18.24 For the autoionization of water at...Ch. 18 - Prob. 18.25QPCh. 18 - Prob. 18.26QPCh. 18 - Prob. 18.27QPCh. 18 - Prob. 18.28QPCh. 18 - Prob. 18.29QPCh. 18 - Prob. 18.30QPCh. 18 - Prob. 18.31QPCh. 18 - Prob. 18.32QPCh. 18 - Prob. 18.33QPCh. 18 - Prob. 18.34QPCh. 18 - Prob. 18.35QPCh. 18 - Prob. 18.36QPCh. 18 - Prob. 18.37QPCh. 18 - Prob. 18.38QPCh. 18 - Prob. 18.39QPCh. 18 - Prob. 18.40QPCh. 18 - Prob. 18.41QPCh. 18 - Prob. 18.42QPCh. 18 - Prob. 18.43QPCh. 18 - Prob. 18.44QPCh. 18 - Prob. 18.45QPCh. 18 - Prob. 18.46QPCh. 18 - 18.47 Calculate the equilibrium pressure of CO2...Ch. 18 - Prob. 18.48QPCh. 18 - 18.49 Referring to Problem 18.48, explain why the...Ch. 18 - Prob. 18.50QPCh. 18 - Prob. 18.51QPCh. 18 - Prob. 18.52QPCh. 18 - Prob. 18.53QPCh. 18 - Prob. 18.54QPCh. 18 - Prob. 18.55QPCh. 18 - 18.56 Crystallization of sodium acetate from a...Ch. 18 - Prob. 18.57QPCh. 18 - Prob. 18.58QPCh. 18 - Prob. 18.59QPCh. 18 - Prob. 18.60QPCh. 18 - Prob. 18.61QPCh. 18 - Prob. 18.62QPCh. 18 - Prob. 18.63QPCh. 18 - Prob. 18.64QPCh. 18 - Prob. 18.65QPCh. 18 - Prob. 18.66QPCh. 18 - Prob. 18.67QPCh. 18 - Prob. 18.68QPCh. 18 - Prob. 18.69QPCh. 18 - Prob. 18.70QPCh. 18 - Prob. 18.71QPCh. 18 - Prob. 18.72QPCh. 18 - 18.73 (a) Over the years there have been numerous...Ch. 18 - Prob. 18.74QPCh. 18 - 18.75 Shown here are the thermodynamic data for...Ch. 18 - Prob. 18.76QPCh. 18 - Prob. 18.77QPCh. 18 - Prob. 18.78QPCh. 18 - Prob. 18.79QPCh. 18 - Prob. 18.80QPCh. 18 - Prob. 18.81QPCh. 18 - Prob. 18.82QPCh. 18 - Prob. 18.83QPCh. 18 - 18.84 Large quantities of hydrogen are needed for...Ch. 18 - Prob. 18.85QPCh. 18 - Prob. 18.86QPCh. 18 - Prob. 18.87QPCh. 18 - Prob. 18.88QPCh. 18 - Prob. 18.89QPCh. 18 - Prob. 18.90QPCh. 18 - Prob. 18.91QPCh. 18 - Prob. 18.92QPCh. 18 - Prob. 18.93QPCh. 18 - Prob. 18.94QPCh. 18 - Prob. 18.95QPCh. 18 - Prob. 18.96QPCh. 18 - Prob. 18.98QPCh. 18 - Prob. 18.100SPCh. 18 - Prob. 18.101SPCh. 18 - Prob. 18.102SPCh. 18 - Prob. 18.103SPCh. 18 - Prob. 18.104SPCh. 18 - Prob. 18.105SPCh. 18 - Prob. 18.106SPCh. 18 - Prob. 18.107SPCh. 18 - Prob. 18.108SPCh. 18 - 18.109 The boiling point of diethyl ether is...
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