Chapter 15.5, Problem 15.12WE

For the same reaction and temperature as in Worked Example 15.11, calculate the equilibrium concentrations of all three species if the starting concentrations are as follows: [H₂] = 0.00623 M, [I₂] = 0.00414 M, and [HI] = 0.0424 M.

Interpretation Introduction

Interpretation:

The HI equilibrium constant (Kc) values should be calculated given the respective molar concentration of reactants and products at ${\text{430}}^{\text{0}}\text{C}$.

Concept Introduction:

Equilibrium constant (K): 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 an 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.

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.

Answer to Problem 15.12WE

The equilibrium concentration (Kc) value is given the hydrogen iodide chemical reaction is showed below.

$\begin{array}{l}{\text{H}}_{\text{2}}\text{(g)}{\text{+I}}_{\text{2}}\text{(g)}\rightleftharpoons 2\text{HI(g)}\\ {\text{K}}_c=\frac{{[HI]}^2}{[H_2][I_2]}\frac{[Product]}{[Reac\tan t]}=54.3\\ \text{The equlibrium concentration of HI}=54.3\end{array}$

Explanation of Solution

To find: Calculate the each $(Kc)$ values for given the hydrogen iodide (HI) equilibrium constant.

First we calculate the equilibrium table to determine the equilibrium concentrations of each species in terms of unknown (x) then solve for (x) and use it to calculate the equilibrium molar concentrations.

Let us consider the given equilibrium concentration values, to substitute (Kc) equation we get the equilibrium constant values.

$\begin{array}{l}{\text{H}}_{\text{2}}\text{(g)}{\text{+I}}_{\text{2}}\text{(g)}\rightleftharpoons 2\text{HI(g)}\\ {\text{K}}_c=\frac{{[HI]}^2}{[H_2][I_2]}\frac{{(0.0424)}^2}{(0.00623)(0.00414)}=69.7\\ \text{The equlibrium constant value HI}=69.7\end{array}$

$\begin{array}{l}{\text{H}}_{\text{2}}\text{(g)}\text{+}{\text{I}}_{\text{2}}\text{(g)}\rightleftharpoons 2\text{HI(g)}\\ \text{Initial (M): }0.006230.006230.0424\\ \text{Change (M): +x}+\text{x}-\text{2x}\\ ----------------------------\\ \text{Eqilibrium (M):}(0.240+x)(0.240+x)(-2x)\\ \text{Here }\\ \text{Given the equilbrium constant (Kc) value is 54}\text{.3}\text{at}\text{430}°\text{C}\\ Theequlibriumcons\tan t(Kc)followedby\\ {\text{K}}_c=\frac{{\text{[HI]}}^{\text{2}}}{[H_2][I_2]}------[1]\\ \text{Calculate}\text{the}\text{equlibrium}\text{value}\text{of}(\text{Kc)}\\ \text{54}\text{.3=}\frac{{(0.0424-2x)}^2}{(0.00623+x)(0.00623+x)}=\frac{{(0.0424-2x)}^{\cancel{2}}}{{(3.88\times {10}^3+x)}^{\cancel{2}}}\\ \text{Thaking}\text{for}\text{squre}\text{root}\text{of}\text{both}\text{side}\text{above}\text{obtained}\text{values}\\ 54.3=\frac{(0.0424-2x)}{(3.88\times {10}^3+x)}\\ \sqrt{54.3}=\frac{(0.0424-2x)}{(3.88\times {10}^3+x)};7.37=\frac{(0.0424-2x)}{(3.88\times {10}^3+x)}\\ \text{Here the respactive concentrations are }{\text{(H}}_{\text{2}}\text{and}{\text{I}}_{\text{2}}\text{) are equal }\\ \text{Instead}\text{to carry out the given muotiplications}\text{.}\\ \text{54}\text{.3(2}\text{.58}\times {\text{10}}^{\text{-5}}\text{+}\text{1}\text{.04}\times {\text{10}}^{\text{-2}}\text{x}\times x^2\text{)=}\text{1}\text{.80}\times {10}^{-3}-1.70\times {10}^{-1}x+4x^2\\ \text{Further we collecting}\text{terms we get }\\ \text{50}{\text{.3x}}^{\text{2}}+0.735x-4.00\times {10}^{-4}=0\end{array}$

$\begin{array}{l}\text{50}{\text{.3x}}^{\text{2}}+0.735x-4.00\times {10}^{-4}=0\\ {\text{We consider the quadradic equation of the form ax}}^{\text{2}}\text{+bx+c=0}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \text{Here }\\ \text{a=50}\text{.3, b=0}\text{.735,}\text{c=}-4.00\times {10}^{-4}\\ x=\frac{-\text{0}\text{.735}\pm \sqrt{{(\text{0}\text{.735})}^2-4(\text{50}\text{.3})(-4.00\times {10}^{-4})}}{2(\text{50}\text{.3})}\\ x=\frac{-\text{0}\text{.735}\pm \sqrt{\text{0}\text{.54022}-201.2(-4.00\times {10}^{-4})}}{100.6}\\ x=\frac{-\text{0}\text{.735}\pm \sqrt{\text{0}\text{.54022+0}\text{.8048}}}{100.6}\\ x=\frac{-\text{0}\text{.735}\pm \sqrt{1.34502}}{100.6}=x=\frac{-\text{0}\text{.735}\pm 1.15974}{100.6}\\ \text{Solve the above values we get,}\\ \text{x=5}\text{.25}\times {\text{10}}^{\text{-4}}(or)x=-0.0151\end{array}$

The equilibrium concentration values are

$\begin{array}{l}\text{Hence the first of these values 5}\text{.25}\times {\text{10}}^{\text{-4}}\text{and calculate value of (x)}\\ {\text{[H}}_{\text{2}}\text{]=}(0.00623-x)M(\text{Here}\text{x=5}\text{.25}\times {\text{10}}^{\text{-4}})\\ {\text{[H}}_{\text{2}}\text{]=}(0.00414-\text{5}\text{.25}\times {\text{10}}^{\text{-4}})M=0.00676M\\ {\text{[I}}_{\text{2}}\text{]=}(0.00414-\text{5}\text{.25}\times {\text{10}}^{\text{-4}})M=0.00467M\\ \text{[HI]=(0}\text{.0424-2x)M}=\text{(0}\text{.0424-2(5}\text{.25}\times {\text{10}}^{\text{-4}}\text{))=}0.0414M\\ \text{The respactive reactants and}\text{products values}\text{are substituted (Kc)}\\ equation.\\ \Rightarrow \frac{{[HI]}^2}{[H_2][I_2]}=\frac{{(0.0414)}^2}{(0.00676)(0.00467)}=54.3\\ \text{Hence Kc=54}\text{.3}\end{array}$

The equal moles of H₂ and I₂ reacted in gas phase conditions to give 2 moles of HI, the balance equation are showed above. Then the depletion in the concentration of HI at equilibrium, the equilibrium concentration of H₂ and I₂ consider as (x), the calculation methods showed the table.

Conclusion

The HI molar concentration is derived given the equilibrium concentration (Kc) and its reaction.