General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 15, Problem 15.72QP
Interpretation Introduction

Interpretation:

The molar concentration (Kc) should be calculated given SO2Cl2 decomposition equilibrium reaction at 648K.

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.

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.

Thermal decomposition reaction: This reaction caused by heat or decomposition of starting substance is the temperature at which the substance chemically decomposes. In other words large molecules being broken down into single elements (or) compounds.

Expert Solution & Answer
Check Mark

Answer to Problem 15.72QP

The equilibrium constant (Kp) values are given the statement of decomposition reaction is presented below.

    SO2Cl2(g)SO2(g)+Cl2(g)Kc=[Product][Reactant]=3.84×102M

Explanation of Solution

To find: The equilibrium reaction should be identified given the statement.

Analyze the chemical equilibrium reaction.

    a).SO2Cl2(g)SO2(g)+Cl2(g)[DessociationReaction]b).SO2(g)+Cl2(g)SO2Cl2(g)[ForamtionReaction]

The given equilibrium concentration reaction is the combined reaction is the product of the constants for this component reaction. This equilibrium reaction expression contains different conditions like solid phase into gases phase, so this process heterogeneous equilibrium the equilibrium constant can also be represented by Kp, were the Kp represents partial pressure. Then the product molecule partial pressure (SO2andCl2) is derived in next method.

To find: Calculate the equilibrium constant values (Kp) are given the SO2Cl2 decomposition reactions.

Calculate the starting components equilibrium constant (Kp) values.

Let us consider the following equilibrium equation.

  a).SO2Cl2(g)SO2(g)+Cl2(g)[DessociationReaction]b).SO2(g)+Cl2(g)SO2Cl2(g)[ForamtionReaction]Weconsidergivendecompositionreaction(a)FormulaweightofSO2Cl2=134.69g/mlHere minimize rounding errors, the initial molarity of SO2Cl2SO2Cl2=Total weight of compound ×formulaweightofcompoundRespactive volumeSO2Cl2=6.75gof SO2Cl2×1molof SO2Cl2135.0gofSO2Cl22.00L=6.75gof SO2Cl2×0.0074072.00=0.049992.00=0.02499(or)0.02500The concentration of SO2at equilibrium isSO2=0.0354mol2.00L=0.01725M

Given the decomposition reaction (1:1) mole ratio between SO2 and SO2Cl2 the concentration of SO2 at equilibrium (0.01725 M) equal the concentration of SO2Cl2 reacted, The ICE concentration equilibrium derived below.

  a).SO2Cl2(g)SO2(g)+Cl2(g)[DessociationReaction]b).SO2(g)+Cl2(g)SO2Cl2(g)[ForamtionReaction]SO2Cl2(g)SO2(g)+Cl2(g)Initial (M): 0.0250000Change (M):  -0.01725+0.01725+0.01725Eqilibrium (M):0.007750.017250.01725Substitutedthe equilibrium concentration into the equilibrium expression to Kc=[Product][Reactant]=[SO2][Cl2][SO2Cl2]=(0.01725)(0.01725)0.00775=0.0002970.00775=0.03849(or)3.84×102

The given decomposition reaction the respective reactant to give products all exists in the different phase  and this equilibrium reaction expression contains single conditions like gases phase, the equilibrium constant can also be represented by Kp and Kc, were the “P” partial pressure. The Kc derived equation showed above.

Conclusion

The molar constant (Kc) values are derived and respective SO2Cl2 decomposition reaction given the statement of equilibrium reaction.

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Chapter 15 Solutions

General Chemistry

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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