Question

Using \(\mathrm{CH}_{4}\) and steam as a source of \(\mathrm{H}_{2}\) for \(\mathrm{NH}_{3}\) synthesis requires high temperatures. Rather than burning \(\mathrm{CH}_{4}\) separately to heat the mixture, it is more efficient to inject some \(\mathrm{O}_{2}\) into the reaction mixture. All of the \(\mathrm{H}_{2}\) is thus released for the synthesis, and the heat of reaction for the combustion of \(\mathrm{CH}_{4}\) helps maintain the required temperature. Imagine the reaction occurring in two steps: $$ \begin{array}{r} 2 \mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+4 \mathrm{H}_{2}(g) \\ K_{\mathrm{p}}=9.34 \times 10^{28} \mathrm{at} 1000 . \mathrm{K} \\ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \quad K_{\mathrm{p}}=1.374 \text { at } 1000 . \mathrm{K} \end{array} $$ (a) Write the overall equation for the reaction of methane, steam, and oxygen to form carbon dioxide and hydrogen. (b) What is \(K_{\mathrm{p}}\) for the overall reaction? (c) What is \(K_{c}\) for the overall reaction? (d) A mixture of \(2.0 \mathrm{~mol}\) of \(\mathrm{CH}_{4}, 1.0 \mathrm{~mol}\) of \(\mathrm{O}_{2},\) and \(2.0 \mathrm{~mol}\) of steam with a total pressure of \(30 .\) atm reacts at \(1000 . \mathrm{K}\) at constant volume. Assuming that the reaction is complete and the ideal gas law is a valid approximation, what is the final pressure?

Short answer

Overall Equation: \(2 \mathrm{CH}_4 + \mathrm{O}_2 + 2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{CO}_2 + 5 \mathrm{H}_2\).\(K_{\mathrm{p}} = 1.283 \times 10^{29}\).\(K_{\mathrm{c}} = 2.32 \times 10^{19}\).Final Pressure: 42 atm.

Step-by-step solution

Write the Overall Equation

Combine the two given reactions to write the overall equation. The first reaction is: 1. \(2 \mathrm{CH}_4(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{CO}(g) + 4 \mathrm{H}_2(g) \)The second reaction is: 2. \(\mathrm{CO}(g) + \mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_2(g) + \mathrm{H}_2(g)\)To find the overall equation, add these two reactions: Overall: \(2 \mathrm{CH}_4(g) + \mathrm{O}_2(g) + 2 \mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons 2 \mathrm{CO}_2(g) + 5 \mathrm{H}_2(g) \).

Determine the Overall Equilibrium Constant \(K_{\mathrm{p}}\)

The equilibrium constants are multiplied when adding reactions: \(K_1 = 9.34 \times 10^{28} \) and \(K_2 = 1.374\), therefore, \(K_{\mathrm{p}} = K_1 \times K_2 = (9.34 \times 10^{28}) \times (1.374)\). Hence, \(K_{\mathrm{p}} = 1.283 \times 10^{29} \).

Calculate the \(K_{\mathrm{c}}\) for the Overall Reaction

Use the relationship between \(K_{\mathrm{p}}\) and \(K_{\mathrm{c}}\): \(K_{\mathrm{p}} = K_{\mathrm{c}} (RT)^{\Delta n}\)Where \(\Delta n\) is the difference in moles of gas products and reactants. \(\Delta n = (2 + 5) - (2 + 1 + 2) = 4\)Using \(R = 0.0821 \;\mathrm{L.atm/mol.K}\) and \(T = 1000\; \mathrm{K}\): \(1.283 \times 10^{29} = K_{\mathrm{c}} (0.0821 \times 1000)^4\)Solving for \(K_{\mathrm{c}}\): \(K_{\mathrm{c}} = 1.283 \times 10^{29} / (82.1)^4 = 2.32 \times 10^{19}\).

Calculate the Final Pressure

Use the ideal gas law and stoichiometry. Initially, we have: \(2.0 \;\mathrm{moles \;CH_4} + 1.0 \; \mathrm{mole \;O_2 } + 2.0 \; \mathrm{moles \;H_2O} = 5.0 \; \mathrm{moles}\). After reaction, we have: \(2 \mathrm{CO_2} + 5 \mathrm{H_2} = 7 \mathrm{moles}\). Using the formula \(P = \frac{nRT}{V}\), pressure is directly proportional to moles in constant volume and temperature.Initially: \(P_i = 30 \;\mathrm{atm}\).Final pressure proportion: \(P_f = 30 \;\mathrm{atm} \times \frac{7 \;\mathrm{moles}}{5 \;\mathrm{moles}} = 42 \;\mathrm{atm}\).

Key concepts

equilibrium constants

A critical component of chemical equilibrium calculations is understanding equilibrium constants. These constants are denoted by symbols like \(\) and \(\), representing the equilibrium constant in terms of partial pressures and concentrations, respectively.

The values of these constants reflect the ratio of the products to reactants at equilibrium under a given set of conditions:

• For a reaction at equilibrium: \(aA + bB \rightleftharpoons cC + dD\) the equilibrium constant based on partial pressures \(K_{\mathrm{p}}\) is defined as \[ K_{\mathrm{p}} = \frac{{P_C^c \cdot P_D^d}}{{P_A^a \cdot P_B^b}} \] where \(P_x\) represents the partial pressure of gas x.
• When using concentrations, the equilibrium constant \(K_{\mathrm{c}}\) is written as: \[ K_{\mathrm{c}} = \frac{{[C]^c \cdot [D]^d}}{{[A]^a \cdot [B]^b}} \]

To relate \(K_{\mathrm{p}}\) and \(K_{\mathrm{c}}\), the equation \(K_{\mathrm{p}} = K_{\mathrm{c}} (RT)^ {\Delta n}\), where \(R\) is the gas constant (0.0821 L.atm/mol.K), \(T\) is the temperature in Kelvin, and \( \Delta n\) is the change in the number of moles of gas (\text{products minus reactants}), is used. This relation is essential in solving equilibrium problems, as evidenced in the practice exercise. For instance, to find \(K_{\mathrm{p}}\) for the overall reaction of methane and steam, we multiply the individual constants of each elementary reaction.

reaction stoichiometry

Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It ensures the conservation of mass by balancing the amounts of substances participating in a reaction.

For the reactions from the exercise:

• \(2 \text{CH}_4(g) + \text{O}_2(g) \rightleftharpoons 2 \text{CO}(g) + 4 \text{H}_2(g)\)
• \(\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)\)

The overall reaction can be obtained by adding the equations: \(2 \text{CH}_4(g) + \text{O}_2(g) + 2 \text{H}_2\text{O}(g) \rightleftharpoons 2 \text{CO}_2(g) + 5 \text{H}_2(g)\).
Combining reactions in stoichiometry often requires understanding their respective equilibrium constants. Balancing the equations ensures that all atoms are accounted for in the transformation from reactants to products.Combining reactions and calculations like those in the workout use these principles to arrive at proper representations of equilibrium states.

ideal gas law

The ideal gas law describes the behavior of ideal gases and is represented by the equation \(PV = nRT\), where:

• \(P\) is the pressure,
• \(V\) is the volume,
• \(n\) is the number of moles,
• \(R\) is the gas constant (0.0821 L.atm/mol.K),
• \(T\) is the temperature in Kelvin.

It is used to relate the physical variables of gases. In the provided exercise, we use the ideal gas law to determine the final pressure of the reacting gases at constant volume and given temperature.

Initially, the total pressure is 30 atm, and the molar quantities of reactants are given. After the reaction completes, the moles of gas products are determined:

• \(\text{Initial: } 2.0 \text{ mol } \text{CH}_4, 1.0 \text{ mol } \text{O}_2, 2.0 \text{ mol } \text{H}_2\text{O} = 5.0 \text{ mol}\)
• \(\text{Final: } 2 \text{CO}_2 + 5 \text{H}_2 = 7 \text{ mol}\)

Using the proportionality of pressure to the number of moles, the final pressure can be determined: \[P_f = 30 \text{ atm} \times \frac{7 \text{ mol}}{5 \text{ mol}} = 42 \text{ atm} \].

thermodynamics

Thermodynamics is the study of energy changes, particularly the heat and work involved in chemical reactions. It plays a vital role in understanding reaction spontaneity and equilibrium. Some key principles include:

• First Law of Thermodynamics: Energy cannot be created or destroyed, only transferred or transformed.
• Second Law of Thermodynamics: In any spontaneous process, the total entropy of the system and surroundings always increases.
• Third Law of Thermodynamics: As temperature approaches absolute zero, the entropy of a perfect crystal approaches zero.

In the provided exercise, thermodynamic principles come into play primarily in the heat management of the reactions. The reaction involving \(\text{CH}_4\) and \(\text{O}_2\) not only produces \(\text{H}_2\) but also generates heat, which assists in maintaining the high temperatures required for further reactions. Thermodynamic calculations often involve enthalpies (heat change at constant pressure), entropies (disorder), and Gibbs free energy (\(\Delta G = \Delta H - T \Delta S\)) to predict reaction feasibility and equilibrium states.