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The equilibrium constant \(K_{P}\) for the reaction $$2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$$is found to be \(2 \times 10^{-42}\) at \(25^{\circ} \mathrm{C}\). (a) What is \(K_{\mathrm{c}}\) for the reaction at the same temperature? (b) The very small value of \(K_{P}\) (and \(K_{\mathrm{c}}\) ) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.

Short Answer

Expert verified
(a) \( K_c \approx 8.23 \times 10^{-44} \). (b) Reaction kinetics prevent spontaneous combustion.

Step by step solution

01

Understand the relationship between Kp and Kc

The relationship between the equilibrium constants \( K_P \) and \( K_c \) for a gaseous reaction is given by the equation \( K_P = K_c (RT)^{\Delta n} \), where \( R \) is the universal 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 (products - reactants).
02

Determine Δn

For the reaction \( 2\text{H}_2\text{O}(g) \rightleftharpoons 2\text{H}_2(g) + \text{O}_2(g) \), \( \Delta n \) is the difference in moles of gaseous products and reactants. Here, \( \Delta n = (2 + 1) - 2 = 1 \).
03

Convert Temperature to Kelvin

Convert the temperature given in Celsius to Kelvin. Use the formula \( T(\text{K}) = T(^{\circ}\text{C}) + 273.15 \). Thus, \( 25^{\circ}\text{C} = 298.15 \text{K} \).
04

Calculate Kc from Kp

Substitute \( K_P = 2 \times 10^{-42} \), \( R = 0.0821 \text{ L·atm/mol·K} \), \( T = 298.15 \text{K} \), and \( \Delta n = 1 \) into the equation \( K_P = K_c (RT)^{\Delta n} \). Solve for \( K_c \):\[ K_c = \frac{K_P}{(RT)^1} = \frac{2 \times 10^{-42}}{(0.0821 \times 298.15)} \approx 8.23 \times 10^{-44} \].
05

Discuss Thermodynamic Stability

The very small value of \( K_P \) and hence \( K_c \) indicates the reaction strongly favors water formation, meaning the reverse reaction is non-spontaneous at room temperature. Consequently, a mixture of hydrogen and oxygen can be kept without reaction because the energy required to initiate the conversion back to water (activation energy) is not surpassed under normal conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The equilibrium constant is a vital concept in understanding chemical equilibrium. It is denoted by different symbols depending on the nature of the substances involved, such as \( K_c \) for concentrations and \( K_P \) for partial pressures in gaseous reactions.

In general, the equilibrium constant gives us an idea of the extent of a reaction. A very large \( K \) value indicates that the products are favored at equilibrium, while a very small \( K \) value shows that the reactants are more stable.

For the reaction \( 2 \mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_2(g) + \mathrm{O}_2(g) \), the equilibrium constant \( K_P \) is given as \( 2 \times 10^{-42} \) at 25°C. This tiny value suggests an overwhelmingly higher concentration of water compared to hydrogen and oxygen at equilibrium, meaning the formation of water is highly favored under these conditions.
  • \( K_P = K_c (RT)^{\Delta n} \)
  • \( \Delta n \) changes based on the difference in moles between products and reactants.
Gaseous Reaction
Gaseous reactions often have unique characteristics due to the behavior of gases under different conditions. They can be expressed in terms of partial pressures, leading to the equilibrium constant \( K_P \).

In the given reaction \( 2 \mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_2(g) + \mathrm{O}_2(g) \), \( \Delta n \) is calculated as the change in moles of gaseous substances from reactants to products. Here, \( \Delta n = 1 \), because the number of moles shifts from 2 (reactants) to 3 (products).

When calculating \( K_P \) from \( K_c \) (or vice versa), it is important to factor in the temperature in Kelvin and the ideal gas constant, R. Observing these reactions helps understand phenomena like reversibility and reaction direction under varying conditions.
  • Temperature converts using \( T(\text{K}) = T(^{\circ}\text{C}) + 273.15 \).
  • \( K_P \) and \( K_c \) express the equilibrium state in terms of pressure and concentration respectively.
Thermodynamic Stability
Thermodynamic stability refers to the energy state of a chemical system; it dictates how likely substances are to persist over possible transformations. In equilibrium scenarios, thermodynamically stable substances are those that do not readily transform without an external input of energy.

With a very small \( K_P \) or \( K_c \), the reaction system is strongly aligned towards one direction—in this case, water formation. Although hydrogen and oxygen are reactive, being able to "stay together" at room temperature shows they are in a low-energy, stable state concerning spontaneous reaction back to water.

The reason why a mixture of hydrogen and oxygen doesn't spontaneously revert to water under normal conditions is due to high activation energy needs. They require considerable energy input to initiate the reaction, such as a spark or added heat. Hence, despite their noted chemistry, they remain inert without such stimuli.
  • Thermodynamically stable species resist change without energy input.
  • Activation energy dictates the likelihood of a reaction occurring spontaneously.

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Most popular questions from this chapter

A mixture containing 3.9 moles of \(\mathrm{NO}\) and 0.88 mole of \(\mathrm{CO}_{2}\) was allowed to react in a flask at a certain temperature according to the equation $$\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{CO}(g)$$At equilibrium, 0.11 mole of \(\mathrm{CO}_{2}\) was present. Calculate the equilibrium constant \(K_{\mathrm{c}}\) of this reaction.

What is meant by "the position of an equilibrium"? Does the addition of a catalyst have any effects on the position of an equilibrium?

Consider the potential energy diagrams for two types of reactions \(\mathrm{A} \rightleftharpoons \mathrm{B}\). In each case, answer the following questions for the system at equilibrium. (a) How would a catalyst affect the forward and reverse rates of the reaction? (b) How would a catalyst affect the energies of the reactant and product? (c) How would an increase in temperature affect the equilibrium constant? (d) If the only effect of a catalyst is to lower the activation energies for the forward and reverse reactions, show that the equilibrium constant remains unchanged if a catalyst is added to the reacting mixture.

A quantity of 1.0 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) was introduced into an evacuated vessel and allowed to attain equilibrium at a certain temperature $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ The average molar mass of the reacting mixture was \(70.6 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of the gases. (b) Calculate \(K_{P}\) for the reaction if the total pressure was 1.2 atm. (c) What would be the mole fractions if the pressure were increased to 4.0 atm by reducing the volume at the same temperature?

When heated, ammonium carbamate decomposes as $$\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)$$ At a certain temperature the equilibrium pressure of the system is 0.318 atm. Calculate \(K_{P}\) for the reaction.

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