Question

Use the following data to estimate a value of $K_{\mathrm{p}}$ at $1200\mathrm{K}$ for the reaction $2{\mathrm{H}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})\rightleftharpoons 2{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$ $$\begin{array}{l}\text{ C(graphite) }+{\mathrm{CO}}_2(\mathrm{g})\rightleftharpoons 2\mathrm{CO}(\mathrm{g})K_{\mathrm{c}}=0.64\\ {\mathrm{CO}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})\rightleftharpoons \mathrm{CO}(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g})K_{\mathrm{c}}=1.4\\ \text{ C(graphite) }+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})\rightleftharpoons \mathrm{CO}(\mathrm{g})K_{\mathrm{c}}=1\times {10}^8\end{array}$$

Short answer

The value of $K_p$, the equilibrium constant for the reaction $2H_2(g)+O_2(g)\rightleftharpoons 2H_{2}O(g)$ at 1200 K is $(1//64)\ast 1.4\ast (1\times {10}^8{)}^2$.

Step-by-step solution

Identifying the Needed Reactions

Inspect given reactions and identify which ones can be manipulated algebraically (through reverse, multiply or addition/subtraction) to obtain the desired reaction $2H_2(g)+O_2(g)\rightleftharpoons 2H_{2}O(g)$. The ones to be used are: \n \n C(graphite) + CO2(g) \rightleftharpoons 2CO(g) with $K_c=0.64$; \n CO2(g) + H2(g) \rightleftharpoons CO(g) + H2O(g) with $K_c=1.4$; \n C(graphite) + 0.5O2(g) \rightleftharpoons CO(g) with $K_c=1\times {10}^8$

Manipulate the Reactions and the Corresponding Equilibrium Constants

In order to add the reactions to achieve the desired final reaction, reverse the first reaction:2CO(g) \rightleftharpoons C(graphite) + CO2(g) with $K_c=1/0.64$,keep the second reaction as is and multiply the third reaction by 2:2C(graphite) + O2(g) \rightleftharpoons 2CO(g) with $K_c=(1\times {10}^8{)}^2$. Since reversing a reaction inverts the equilibrium constant and changing the coefficient of the reaction elevates the $K_c$ to that power.

Adding up Reactions and Corresponding Equilibrium Constants

Add up the reactions manipulated in step one:$$2CO(g)\rightleftharpoons C(graphite)+CO2(g)$$$$CO2(g)+H2(g)\rightleftharpoons CO(g)+H2O(g)$$$$2C(graphite)+O2(g)\rightleftharpoons 2CO(g)$$This results in the desired final reaction: $$2H_2(g)+O2(g)\rightleftharpoons 2H_{2}O(g)$$In the same manner, add up the corresponding $K_c$s by multiplying them together, as when adding up reactions, the corresponding equilibrium constants multiply. Therefore, the final $K_c$ for the desired reactions is: $(1/0.64)\ast 1.4\ast (1\times {10}^8{)}^2$.

Conversion from $K_c$ to $K_p$

Since the question asks for the value of $K_p$ and the resultant equilibrium constant achieved is $K_c$, transform $K_c$ to $K_p$ using the formula $K_p=K_c(RT{)}^{\mathrm{\Delta }n}$ where $R$ is the ideal gas constant (0.08314 L bar /mol K), $T$ is the absolute temperature in Kelvin (1200 K in this case), and $\mathrm{\Delta }n$ is the change in stoichiometric number (sum of the stoichiometric coefficient of the gaseous products minus the sum of the stoichiometric coefficient of the gaseous reactants). Substituting the obtained $K_c$ and given $T$ value and with the $\mathrm{\Delta }n$ being 0 in this given reaction because the number of gaseous moles remains the same on both sides of the reaction, solve for $K_p$.

Calculate $K_p$

As $\mathrm{\Delta }n$ is zero, the equation simplifies to $K_p=K_c$. The final $K_p$ will be $(1//64)\ast 1.4\ast (1\times {10}^8{)}^2$, after doing the math.

Key concepts

Chemical Equilibrium

Chemical equilibrium refers to a state in a reversible chemical reaction where the rates of the forward and backward reactions are equal. This means there is no net change in the concentration of reactants and products over time.
At this point, the reaction appears to be static, but in reality, the molecules are still dynamically interacting.
It is important to remember that equilibrium means concentrations remain constant, not necessarily equal.
The equilibrium constant ( K ) expresses the ratio of product concentrations to reactant concentrations at equilibrium for a particular reaction.
The value of K indicates the extent of a reaction at equilibrium:

• A large K means that at equilibrium, the reaction mixture contains mostly products.
• A small K suggests that at equilibrium, the mixture contains mainly reactants.

Understanding chemical equilibrium is crucial because it helps predict how a reaction mixture will behave under different conditions.

Temperature Dependence of Equilibrium

The equilibrium constant is significantly influenced by temperature.
Changes in temperature affect chemical reactions by shifting the position of equilibrium.
According to Le Chatelier’s Principle, when the temperature of a reaction is changed, the system adjusts to counteract that change and establishes a new equilibrium.
This leads to either an increase or decrease in the equilibrium constant, depending on the reaction being endothermic or exothermic:

• Endothermic reactions consume heat ( ΔH > 0 ). Increasing the temperature will shift the equilibrium to favor the products, increasing K .
• Exothermic reactions release heat ( ΔH < 0 ). Increasing the temperature shifts equilibrium to favor reactants, decreasing K .

This means that knowing the temperature dependence on K allows us to predict changes in reaction composition when the temperature varies.

Reversible Reactions

Reversible reactions are chemical reactions that can proceed in both the forward and reverse directions.
These reactions do not go to completion; instead, they reach a point where both reactants and products are present.
A well-known example of a reversible reaction is the synthesis of ammonia ( N_2 + 3H_2 ightleftharpoons 2NH_3 ). At equilibrium, both nitrogen and hydrogen are present alongside ammonia.
Understanding reversible reactions is important because:

• They allow for control of reaction conditions to maximize yield of products.
• They demonstrate the dynamic nature of chemical interactions.

Furthermore, the equilibrium concept is closely tied to reversible reactions, as it explains where the balance lies in these dynamic systems.

Gaseous Reactions

Gaseous reactions are chemical processes involving gases as either reactants, products, or both.
In the context of equilibrium, it is essential to consider the gaseous state because pressure and volume can significantly affect the system.
For gaseous reactions, the equilibrium constant can be expressed in terms of pressures ( K_p ) as opposed to concentrations ( K_c ). This is particularly useful when dealing with reactions involving gases, as pressures are often easier to measure:

• K_p and K_c are related by the equation K_p = K_c(RT)^{Δn} , where R is the ideal gas constant and Δn represents the change in moles of gas.

This calculation is crucial for translating concentration-based constants into pressure-based ones, allowing for a complete understanding of gaseous reactions in equilibrium. Consequently, studying gaseous reactions helps predict reaction behavior in real-world applications, like industrial synthesis processes.