Chapter 15, Problem 15.72QP

Interpretation Introduction

Interpretation:

The molar concentration (Kc) should be calculated given ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ decomposition equilibrium reaction at $\text{648K}$.

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.

Answer to Problem 15.72QP

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

    $\begin{array}{l}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{SO}}_{\text{2}}\text{(g)}\text{+}{\text{Cl}}_{\text{2}}\text{(g)}\\ {\text{K}}_c=\frac{[Product]}{[Reac\tan t]}=3.84\times {10}^{-2}\text{M}\end{array}$

Explanation of Solution

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

Analyze the chemical equilibrium reaction.

    $\begin{array}{l}\text{a)}{\text{.SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{SO}}_{\text{2}}\text{(g)}\text{+}{\text{Cl}}_{\text{2}}\text{(g)}----[DessociationReaction]\\ \text{b)}{\text{.SO}}_{\text{2}}\text{(g)}\text{+}{\text{Cl}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}----[Foramtion\text{}Reaction]\end{array}$

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 $\left({\text{SO}}_{\text{2}}\text{and}{\text{Cl}}_{\text{2}}\right)$ is derived in next method.

To find: Calculate the equilibrium constant values (Kp) are given the ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ decomposition reactions.

Calculate the starting components equilibrium constant (Kp) values.

Let us consider the following equilibrium equation.

  $\begin{array}{l}\text{a)}{\text{.SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{SO}}_{\text{2}}\text{(g)}\text{+}{\text{Cl}}_{\text{2}}\text{(g)}----[DessociationReaction]\\ \text{b)}{\text{.SO}}_{\text{2}}\text{(g)}\text{+}{\text{Cl}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}----[Foramtion\text{}Reaction]\\ Weconsidergiven\text{}decompositionreaction(a)\\ Formulaweightof{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}=134.69g/ml\\ {\text{Here minimize rounding errors, the initial molarity of SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\\ {\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}=\frac{\text{Total weight of compound }\times \text{formula}\text{weight}\text{of}\text{compound}}{\text{Respactive volume}}\\ {\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{=}\frac{\text{6}\text{.75}\text{g}{\text{of SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}{\text{of SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}{\text{135}\text{.0}\text{g}{\text{ofSO}}_{\text{2}}{\text{Cl}}_{\text{2}}}}{\text{2}\text{.00}\text{L}}\\ =\frac{\text{6}\text{.75}\text{g}{\text{of SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\times 0.007407}{2.00}=\frac{0.04999}{2.00}=0.02499(or)0.02500\\ {\text{The concentration of SO}}_{\text{2}}\text{at equilibrium is}\\ {\text{SO}}_{\text{2}}=\frac{0.0354mol}{2.00L}=0.01725M\end{array}$

Given the decomposition reaction (1:1) mole ratio between ${\text{SO}}_{\text{2}}$ and ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ the concentration of ${\text{SO}}_{\text{2}}$ at equilibrium (0.01725 M) equal the concentration of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ reacted, The ICE concentration equilibrium derived below.

  $\begin{array}{l}\text{a)}{\text{.SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{SO}}_{\text{2}}\text{(g)}\text{+}{\text{Cl}}_{\text{2}}\text{(g)}----[DessociationReaction]\\ \text{b)}{\text{.SO}}_{\text{2}}\text{(g)}\text{+}{\text{Cl}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}----[Foramtion\text{}Reaction]\\ {\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{SO}}_{\text{2}}\text{(g)}\text{+}{\text{Cl}}_{\text{2}}\text{(g)}\\ \text{Initial (M): }0.02500\text{0}\text{0}\\ \text{Change (M): -0}\text{.01725}+\text{0}\text{.01725}+\text{0}\text{.01725}\\ ----------------------------\\ \text{Eqilibrium (M):}0.007750.017250.01725\\ \text{Substituted}\text{the equilibrium concentration into the equilibrium expression to }\\ {\text{K}}_c=\frac{[Product]}{[Reac\tan t]}=\frac{{\text{[SO}}_{\text{2}}][{\text{Cl}}_{\text{2}}\text{]}}{[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}]}=\frac{(0.01725)(0.01725)}{0.00775}\\ =\frac{0.000297}{0.00775}=0.03849(or)3.84\times {10}^{-2}\end{array}$

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 ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ decomposition reaction given the statement of equilibrium reaction.