Chapter 15, Problem 15.62QP

Consider the following equilibrium process at 686°C.

CO₂(g) + H₂(g) ⇄ CO(g) + H₂O(g)

The equilibrium concentrations of the reacting species are [CO] = 0.050 M, [H₂] = 0.045 M, [CO₂] = 0.086 M, and [H₂O] = 0.040 M.

(a) Calculate K_C for the reaction at 686°C.

(b) If we add CO₂ to increase its concentration to 0.50 mol/L, what will the concentrations of all the gases be when equilibrium is reestablished?

Interpretation Introduction

Interpretation:

The equilibrium concentration (Kc) should be calculated to given the statement of equilibrium reaction at ${686}^{\text{ο}}\text{C}$.

Concept Introduction:

Chemical equilibrium: 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.

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.

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.

Answer to Problem 15.62QP

The equilibrium concentration (Kc) value is given the statement of chemical reaction is showed below.

$\begin{array}{l}{\text{CO}}_{\text{2}}\text{(g)}+{\text{H}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{CO}}_{\text{2}}\text{(g)}+{\text{H}}_{\text{2}}\text{O(g)}\\ {\text{K}}_c=\frac{[H_{2}O][C{O}_2]}{[C{O}_2][H_2]}=\frac{[Product]}{[Reac\tan t]}=0.52\\ Qc=0.089\\ \text{The}\text{respactive}\text{molar}\text{values}\\ 0.48,0.020,0.075,0.065\end{array}$

Explanation of Solution

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

Calculate and analyze the $(Kc)$ values at 686⁰C.

The consider the following expression of (Kc)

$\begin{array}{l}{\text{CO}}_{\text{2}}\text{(g)}+{\text{H}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{CO}}_{\text{2}}\text{(g)}+{\text{H}}_{\text{2}}\text{O(g)}\\ {\text{K}}_c=\frac{[H_{2}O][C{O}_2]}{[C{O}_2][H_2]}-------[1]\\ \text{The}\text{given}\text{statemnt}\text{of}\text{molar}\text{values}\text{substituted}\text{for}\text{above}\text{equation}\\ =\frac{(0.040)(0.050)}{(0.086)(0.045)}=0.52\end{array}$

The simple equilibrium constant is derived showed above equation (1).

To find: Calculate the molar values for given the statement of equilibrium reaction (b).

Calculate and analyze the molar values at 686⁰C.

First we derived for reaction quotient (Qc) values for given equilibrium reaction.

$\begin{array}{l}{\text{CO}}_{\text{2}}\text{(g)}+{\text{H}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{CO}}_{\text{2}}\text{(g)}+{\text{H}}_{\text{2}}\text{O(g)}\\ {\text{Q}}_{\text{c}}=\frac{[H_{2}O][C{O}_2]}{[C{O}_2][H_2]}\\ {\text{The concentartion of CO}}_{\text{2}}\text{ =0}\text{.50mol/L}\\ \text{This values are substitued for above (Kc) equlibrium reaction}\\ \text{The}\text{given}\text{statemnt}\text{of}\text{molar}\text{values}\text{substituted}\text{for}\text{above}\text{equation}\\ \text{Qc}=\frac{(0.040)(0.050)}{(0.50)(0.045)}=0.089\\ Qc=0.089\end{array}$

Calculate the molar concentration of the given equilibrium reaction.

$\begin{array}{l}{\text{CO}}_{\text{2}}\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{(g)}\rightleftharpoons \text{}\text{CO(g)}{\text{H}}_{\text{2}}\text{O(g)}\\ \text{Initial (M): }\text{0}\text{.50}\text{0}.0450.0500.040\\ \text{Change (M): }\text{}-\text{x}-\text{x}+\text{x}+\text{x}\\ ---------------------------------\\ \text{Eqilibrium (M):}(0.50-x)(0.045-x)(0.050+x)(0.040+x)\\ Theequlibriumcons\tan t(Kc)followedby\\ {\text{K}}_c=\frac{{\text{[H}}_{\text{2}}\text{O][CO]}}{[H_2][C{O}_2]}-=\frac{[Product]}{[Reac\tan t]}------[1]\\ \text{Calculate}\text{the}\text{equlibrium}\text{value}\text{of}(\text{Kc)}\\ \text{Given the respactive molar values }\\ {\text{K}}_c=\frac{{\text{[H}}_{\text{2}}\text{O][CO]}}{[H_2][C{O}_2]}=\frac{(0.040+x)(0.050+x)}{(0.50-x)(0.045-x)}=0.52----[2]\\ \text{Rewrite the equation (2)}\\ \text{0}{\text{.52 (x}}^{\text{2}}\text{-0}\text{.545x}\text{+0}{\text{.0225)=x}}^{\text{2}}\text{+(0}\text{.090x}\text{+}\text{0}\text{.0020)}\\ \text{0}{\text{.48x}}^{\text{2}}+0.373x-(9.7\times {10}^{-3})=0\\ Here\\ x=(0.853)-(9.7\times {10}^{-3})\\ x=0.025\\ \text{The equlibrium concentration are}\\ {\text{Reactant [CO}}_{\text{2}}\text{]=(0}\text{.50-0}\text{.025)M=0}\text{.48M}\\ {\text{[H}}_{\text{2}}\text{]}\text{=}\text{(0}\text{.045-0}\text{.025)M}\text{=0}\text{.020M}\\ \text{Product}[CO]=(0.050+0.025)M=0.075M\\ [H_{2}O]=(0.040+0.025)M=0.065M\end{array}$

The given equilibrium process, the equal moles of H₂ and CO₂ reacted with in gas phase conditions to give a H₂O and CO molecule, the reactant moles 0.50m/L. Then the number of moles calculating (Kc) molar concentration of reacting procedure as fallows, first construct an above equilibrium table and fill in the initial concentration and zero also include the respective table than we use the initial concentration to calculate the reaction quotient (Qr) and compare (K) values to identified the direction in which the reaction will proceeds. Further calculate (x) as the amount of particular species consumed and use the stoichiometry of the reaction to define in terms of (x) the amount of the species produced. Finally the each species in the equilibrium add the change in concentration to the initial concentration to get the equilibrium concentration. The derived molar concentration values are showed above.

Conclusion

The product molecule molar concentration is derived given the equilibrium concentration (Kc) and respective equilibrium reaction at ${686}^{\text{ο}}\text{C}$.