Chapter 15, Problem 15.121QP

The equilibrium constant K_c for the reaction

${\text{H}}_{\text{2}}\left(\text{g}\right)+{\text{I}}_{\text{2}}\left(g\right)\rightleftarrows \text{2HI}\left(g\right)$

is 54.3 at 430°C. At the start of the reaction, there are 0.714 mole of H₂, 0.984 mole of I₂, and 0.886 mole of HI in a 2.40-L reaction chamber. Calculate the concentrations of the gases at equilibrium.

Interpretation Introduction

Interpretation:

Calculate the concentration of gases equilibrium process of given (HI) equilibrium constant (Kc) reaction with respective temperature at ${430}^{\text{ο}}\text{C}$.

Concept Introduction:

Chemical equilibrium reaction: The term applied to reversible chemical reactions. It is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. The equilibrium is achieved; the concentrations of reactant and products become constant.

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.

Homogeneous equilibrium: A homogeneous equilibrium involved has a everything present in the same phase and same conditions, for example reactions where everything is a gas, or everything is present in the same solution.

Answer to Problem 15.121QP

The equilibrium concentration of (Kp and Kc) values for the given gases equilibrium $\text{HI}$ formation reaction is shown below.

$\begin{array}{l}{\text{H}}_{\text{2}}(g)+{\text{I}}_{\text{2}}(g)\rightleftharpoons \text{2HI(g)}\\ \text{Kp=}\frac{{(P_{H{I}_2})}^2}{(P_{H_2})(P_{I_2})}=1.11\\ \text{Kc=}\frac{{[HI]}^2}{[H_2][I_2]}=54.3\\ {\text{H}}_{\text{2}}\text{=0}\text{.070}\text{M}{\text{; I}}_{\text{2}}\text{=0}\text{.182 M}\text{;}\text{HI=0}\text{.825}\text{M}\end{array}$

$\begin{array}{l}{\text{H}}_{\text{2}}(g)+{\text{I}}_{\text{2}}(g)\rightleftharpoons \text{2HI(g)}----[FormationReaction]\\ \text{2HI(g)}\rightleftharpoons {\text{H}}_{\text{2}}(g)+{\text{I}}_{\text{2}}(g)\text{---------[Dessociation Reaction]}\end{array}$

Explanation of Solution

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

Analyze the chemical equilibrium reaction.

The given equilibrium concentration reaction is the combined reaction is the product of the constants for this component reaction. This equilibrium reaction expression contains same conditions like gases phase, so this process homogenous equilibrium the equilibrium constant can also be represented by K and Kcp, were the Kp represents partial pressure and Kc molar constant. Then the each (reactant and product) molecule partial pressure $\left({\text{H}}_{\text{2}}{\text{, I}}_{\text{2}}\text{and HI}\right)$ is derived in step-2.

Interpretation Introduction

Interpretation:

Calculate the concentration of gases equilibrium process of given (HI) equilibrium constant (Kc) reaction with respective temperature at ${430}^{\text{ο}}\text{C}$.

Concept Introduction:

Chemical equilibrium reaction: The term applied to reversible chemical reactions. It is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. The equilibrium is achieved; the concentrations of reactant and products become constant.

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.

Homogeneous equilibrium: A homogeneous equilibrium involved has a everything present in the same phase and same conditions, for example reactions where everything is a gas, or everything is present in the same solution.

Answer to Problem 15.121QP

The equilibrium concentration of (Kp and Kc) values for the given gases equilibrium $\text{HI}$ formation reaction is shown below.

$\begin{array}{l}{\text{H}}_{\text{2}}(g)+{\text{I}}_{\text{2}}(g)\rightleftharpoons \text{2HI(g)}\\ \text{Kp=}\frac{{(P_{H{I}_2})}^2}{(P_{H_2})(P_{I_2})}=1.11\\ \text{Kc=}\frac{{[HI]}^2}{[H_2][I_2]}=54.3\\ {\text{H}}_{\text{2}}\text{=0}\text{.070}\text{M}{\text{; I}}_{\text{2}}\text{=0}\text{.182 M}\text{;}\text{HI=0}\text{.825}\text{M}\end{array}$

$\begin{array}{l}{\text{H}}_{\text{2}}(g)+{\text{I}}_{\text{2}}(g)\rightleftharpoons \text{2HI(g)}----[FormationReaction]\\ \text{2HI(g)}\rightleftharpoons {\text{H}}_{\text{2}}(g)+{\text{I}}_{\text{2}}(g)\text{---------[Dessociation Reaction]}\end{array}$

Explanation of Solution

To find: Calculate the reaction quotient (Qc) values for given the statement of equilibrium reaction.

Calculate and analyze the (Qc) values at ${430}^{\text{0}}\text{C}$.

Let us consider solving of each partial pressure values of given reactions.

$\begin{array}{l}{\text{Given the stating, product moles of H}}_{\text{2}},{\text{I}}_{\text{2}}\text{and}\text{HI}\\ {\text{H}}_{\text{2}}=0.714mol,{\text{I}}_{\text{2}}=0.984mol,\text{HI}=0.886andgivenvolume2.40L\\ \text{Here,}\\ {{\text{[H}}_{\text{2}}}_{\text{]}}\text{=}\frac{0.714mol}{2.40L}=0.298M;\\ {{\text{[I}}_{\text{2}}}_{\text{]}}\text{=}\frac{0.984mol}{2.40L}=0.410M;\\ {\text{[HI]}}_{\text{0}}\text{=}\frac{0.886mol}{2.40L}=0.369M;\\ \text{Calculating}\text{the (Qc)}\text{by substituting the initial concentration }\\ \text{into the approprite below equation (1)}\\ \text{Qc=}\frac{{[HI]}^2}{{[H_2]}_0{[I_2]}_0}-----[1]\\ =\frac{{(0.369)}^2}{(0.298)(0.410)}=1.11\end{array}$

We find the (Qc) is less then Kc. The equilibrium will shifted right side, decreasing the concentrations of (H₂) and (I₂) and increasing concentrations of (HI).

Interpretation Introduction

Interpretation:

Calculate the concentration of gases equilibrium process of given (HI) equilibrium constant (Kc) reaction with respective temperature at ${430}^{\text{ο}}\text{C}$.

Concept Introduction:

Chemical equilibrium reaction: The term applied to reversible chemical reactions. It is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. The equilibrium is achieved; the concentrations of reactant and products become constant.

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.

Homogeneous equilibrium: A homogeneous equilibrium involved has a everything present in the same phase and same conditions, for example reactions where everything is a gas, or everything is present in the same solution.

Answer to Problem 15.121QP

The equilibrium concentration of (Kp and Kc) values for the given gases equilibrium $\text{HI}$ formation reaction is shown below.

$\begin{array}{l}{\text{H}}_{\text{2}}(g)+{\text{I}}_{\text{2}}(g)\rightleftharpoons \text{2HI(g)}\\ \text{Kp=}\frac{{(P_{H{I}_2})}^2}{(P_{H_2})(P_{I_2})}=1.11\\ \text{Kc=}\frac{{[HI]}^2}{[H_2][I_2]}=54.3\\ {\text{H}}_{\text{2}}\text{=0}\text{.070}\text{M}{\text{; I}}_{\text{2}}\text{=0}\text{.182 M}\text{;}\text{HI=0}\text{.825}\text{M}\end{array}$

$\begin{array}{l}{\text{H}}_{\text{2}}(g)+{\text{I}}_{\text{2}}(g)\rightleftharpoons \text{2HI(g)}----[FormationReaction]\\ \text{2HI(g)}\rightleftharpoons {\text{H}}_{\text{2}}(g)+{\text{I}}_{\text{2}}(g)\text{---------[Dessociation Reaction]}\end{array}$

Explanation of Solution

To find: Calculate the equilibrium constant (Kc) values for given the statement of equilibrium reaction.

Calculate and analyze the (Kc) values at ${430}^{\text{0}}\text{C}$.

$\begin{array}{l}\text{Here set up the (ICE) table }\\ {\text{Let (x) be the decrease in concentration of (H}}_{\text{2}}\text{and}{\text{I}}_{\text{2}}\text{)}\\ {\text{H}}_{\text{2}}(g)+{\text{I}}_{\text{2}}(g)\rightleftharpoons 2\text{HI(g)}\\ \text{Initial (M): }0.2980.4100.369\\ \text{Change (M): }\text{-x}-x+2x\\ ----------------------------\\ \text{Eqilibrium (M):}(0.298-x)(0.410-3x)(0.369+2x)\\ \text{The equilibrium expression is }\\ \text{Kc=}\frac{{[HI]}^2}{[H_2][I_2]}-----[2]\\ The(ICE)tablevaluesaresubstitutedequation(2)\\ =\frac{{(0.369+2x)}^2}{(0.298-x)((0.410-3x))}\\ \text{The equilibrium constant for (HI) formation reactions (Kc=54}\text{.3)}\\ 54.3=\frac{{(0.369+2x)}^2}{(0.298-x)((0.410-3x))}\\ \text{We solved a quadratic equation }\\ \frac{-b\pm \sqrt{b^2-4ac}}{2a}-----[3]\\ a=50.3,b=-39.9,c=6.49\\ \text{This values are substituted equation (3)}\\ =\frac{-39.9\pm \sqrt{{(39.9)}^2-4(50.3)(6.49)}}{2(50.3)}\\ =50.3x^2-39.9x+6.49=0\\ \text{Taking}\text{the square root values for above equation}\text{. }\\ \text{x=0}\text{.228}\text{(Here x is the samller)}\\ \text{Hence solved reactant and product equilibrium concentrations are}\\ {\text{[H}}_{\text{2}}\text{]=}\text{(0}\text{.298-0}\text{.228)M}\\ \text{=}\text{0}\text{.070M}\\ {\text{[I}}_{\text{2}}\text{]=}\text{(0}\text{.410-0}\text{.228)M}\\ \text{=0}\text{.182 M}\\ \text{[HI]=}\text{(0}\text{.369)}\text{+}\text{2(0}\text{.228)=}\text{(0}\text{.369)+}\text{(0}\text{.456)}\\ \text{=}\text{0}\text{.825}\text{M}\end{array}$

The each molar concentration values are calculated given the (HI) formation reactions.