Connect 1 Semester Access Card for General Chemistry: The Essential Concepts
Connect 1 Semester Access Card for General Chemistry: The Essential Concepts
7th Edition
ISBN: 9781259692543
Author: Raymond Chang Dr.; Kenneth Goldsby Professor
Publisher: McGraw-Hill Education
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Chapter 15, Problem 15.34QP
Interpretation Introduction

Interpretation:

The equilibrium concentration of iodine gases has to be calculated.

Concept Introduction:

Equilibrium concentration: If Kc and the initial concentration for a reaction and calculate for both equilibrium concentration, and using the (ICE) chart and equilibrium constant and derived changes in respective reactants and products.

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.

Heterogeneous equilibrium: This equilibrium reaction does not depend on the amounts of pure solid and liquid present, in other words heterogeneous equilibrium, substances are in different phases. 

Kp and Kc: This equilibrium constants of gaseous mixtures, these difference between the two constants is that Kc is defined by molar concentrations, whereas Kp is defined by the partial pressures of the gasses inside a closed system.

Vaporized equilibrium: This conversion of liquid in gaseous phase is known as vaporization process. At starting the rate of condensation is less than the rate of evaporation but as evaporation continues the concentration of gaseous molecule in the vapour phase increase.

Expert Solution & Answer
Check Mark

Answer to Problem 15.34QP

The equilibriumconcentrations[I]=8.58×10-4Mand[I2]=0.0194M

Explanation of Solution

To find: The each reactant product equilibrium concentration should be identified given the gas phase reaction.

Write and Analyze the given gas phase chemical equilibrium reaction.

a).I2(g)2I(g)[DessociationReaction]

The given equilibrium reaction has a homogenous process, then the equilibrium constant can also be represented by Kp, were the Kp represents partial pressure. Then the product molecule partial pressure Connect 1 Semester Access Card for General Chemistry: The Essential Concepts, Chapter 15, Problem 15.34QP is derived in step-2.

To find: Calculate equilibrium concentration (Kp) values for given the statement of equilibrium reaction.

Calculate and analyze the (Kp) values at 10000C.

We derived here (Kp) values of (I2) dissociation reaction

First we derived the initial concentration of (I2) is

The initialconcentrationof(I2)=0.0456mol2.30L=0.0198Mhere the 1moleof(I2) dissocoatingto2 molesof (I)atomsLet(x)amount inmol/Lod(I2)dissociatedThe equilibrium concentrationof(I)atoms mustbe=2x

Here set up the (ICE) table Let (x) be the decrease in concentration of  (I2and2I)I2(g)2I(g)Initial (M): 0.01980.000Change (M):  -x+2xEqilibrium (M):(0.0198x)2x

We consider the equilibrium expression in terms of the equilibrium concentration.

The equilibrium constant solvefor(x)Kc=[I]2[I2][1]The(ICE)tablevaluesaresubstitutedequation(1)=(2x)2(0.0198x)=4x2(0.0198x)Given(Kc)valuesare3.80×105Hence,3.80×105=4x2(0.0198x)[2]Rewritetheaboveequation(2)4x2=(3.80×105)x(7.52×107)=0We solved a quadratic equation fromax2+bx+c=0b±b24ac2a[3]a=4,b=3.80×105c=7.52×107This values are substituted equation (3)x=(3.80×105)±(3.80×105)24(4)(7.52×107)2(4)x=(3.80×105)±(3.47×103)8x=4.29×104Mandx=4.29×104M

The obtained second (x) values are negative concentration, this physically impossible so we omitted this values. First (x) value is correct one.

The (x) valuesaresubstituted (ICE) equilibrium values[I]= 2X=(2)(4.29×10-4M)=8.58×10-4M[I2]=(0.0198-x)M=(0.0198-4.29×10-4)=0.0194M

The given iodine dissociation equilibrium reaction the respective reactant to give the two moles of products in the gas phase  and this equilibrium reaction expression contains single conditions like gases phase, the equilibrium constant can also be represented by Kp, were the “P” partial pressure. The each molar concentration values are Kp derived given the gas phase reaction at 10000C as showed above.

Conclusion

The molar concentration (M) values are derived given the iodine (I2) dissociation equilibrium reactions.

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Chapter 15 Solutions

Connect 1 Semester Access Card for General Chemistry: The Essential Concepts

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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