Problem 30 - Chemistry The Molecular Science

Question

Given these data at a certain temperature, $$ \begin{array}{cc} 2 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g}) & K=1.2 \times 10^{-35} \\ \mathrm{~N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) & K=4.6 \times 10^{-3} \\ \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NO}_{2}(\mathrm{~g}) & K=4.1 \times 10^{-9} \end{array} $$ calculate \(K\) for the reaction between 1 mol dinitrogen oxide gas and oxygen gas to give dinitrogen tetraoxide gas.

Step-by-Step Solution

Answer

The equilibrium constant \(K\) for the reaction is approximately \(1.57 \times 10^{23}\).

Step 1: Write the target reaction

The target reaction is the reaction between dinitrogen oxide gas and oxygen to form dinitrogen tetraoxide. We can write the balanced chemical reaction as: \[2 \mathrm{N_2O(g) + O_2(g) \rightarrow N_2O_4(g)}\] Understanding this equation is critical before proceeding further.

Step 2: Use given reactions to form the target reaction

To achieve the desired reaction, we need to combine and manipulate the given reactions. The strategy is to reverse or modify parts of the given equations to match the target reaction. The main reactions provided are: 1. \[2\mathrm{N_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{N_2O(g)}\] with \(K_1 = 1.2 \times 10^{-35}\)2. \[\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}\] with \(K_2 = 4.6 \times 10^{-3}\)3. \[\frac{1}{2}\mathrm{N_2(g)} + \mathrm{O_2(g)} \rightleftharpoons \mathrm{NO_2(g)}\] with \(K_3 = 4.1 \times 10^{-9}\)

Step 3: Manipulate equations to form the target equation

To form the target equation, we need these modifications:- Reverse the first equilibrium reaction to form N2 from N2O:\[2\mathrm{N_2O(g)} \rightarrow 2\mathrm{N_2(g)} + \mathrm{O_2(g)}\] which implies reversing the reaction and using the inverse of the equilibrium constant; thus \(K_1'=\frac{1}{1.2 \times 10^{-35}} = 8.33 \times 10^{34}\).- Use the second equilibrium as it is (since it already produces N2O4).- Take the third reaction as it is.

Step 4: Combine reactions to find the overall equilibrium constant

Combine the equations derived from Step 3:1. \[2\mathrm{N_2O(g)} \rightarrow 2\mathrm{N_2(g)} + \mathrm{O_2(g)}\] with \(K_1' = 8.33\times10^{34}\)2. \[\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}\] with \(K_2 = 4.6 \times 10^{-3}\)3. \[\frac{1}{2}\mathrm{N_2(g)} + \mathrm{O_2(g)} \rightleftharpoons \mathrm{NO_2(g)}\] with \(K_3 = 4.1 \times 10^{-9}\)After reversing the first reaction, the overall equilibrium constant for the desired reaction combining these steps should be calculated by multiplying these constants. This will ultimately be obtainable through careful equation handling without direct interconversion; the manipulations are, however, key for chemical insight.

Step 5: Calculate the combined equilibrium constant

To find the equilibrium constant for the overall reaction, we need to multiply the individual equilibrium constants from the manipulated reactions.- Since we reverse the first reaction, we obtain \(K_1' = 8.33 \times 10^{34}\).- Keep the second reaction as it is with \(K_2 = 4.6 \times 10^{-3}\) and the third unaltered as well. Now, we combine them:\[K_{overall} = (K_1' \times K_2 \times K_3) = \left(8.33 \times 10^{34}\right) \times (4.6 \times 10^{-3}) \times \left(4.1 \times 10^{-9}\right)\]Calculate the resulting \(K\), which corresponds to the overall equilibrium for the reaction.

Step 6: Final calculation and check

After performing the multiplication:\[K_{overall} = (8.33 \times 10^{34}) \times (4.6 \times 10^{-3}) \times (4.1 \times 10^{-9}) \approx 1.57 \times 10^{23} \]This value represents the equilibrium constant for the target reaction: conversion of N2O and O2 into N2O4.

Key Concepts

Reaction Manipulation

In chemistry, manipulating reactions involves adjusting given chemical equations to derive a desired chemical reaction. It is essential for calculating equilibrium constants for complex reactions.
In the given problem, we had to manipulate three separate reactions to obtain the target reaction of

• 2\mathrm{N_2O(g) + O_2(g) \rightarrow N_2O_4(g)}

To achieve this, we reversed the first reaction, which changes the direction of how the chemicals react. Reversing an equation requires taking the reciprocal of its equilibrium constant.
For instance, reversing the reaction:

• 2\mathrm{N_2(g) + O_2(g) \rightleftharpoons 2N_2O(g)}

results in:

• 2\mathrm{N_2O(g) \rightarrow 2N_2(g) + O_2(g)}

with a new equilibrium constant of \(1/1.2 \times 10^{-35} = 8.33 \times 10^{34}\). Reaction manipulation allows chemists to gain insights and form desired products through strategic combination of equations. The core idea involves rearranging given reactions to express a chemical process of interest.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, causing the concentrations of reactants and products to remain constant over time. This state is quantitatively expressed using the equilibrium constant, denoted as \(K\).
The equilibrium constant \(K\) gives insight into the position of equilibrium and is calculated from the concentrations (or pressures) of all reactants and products in a reaction at equilibrium. For example, if a reaction has a large \(K\), the equilibrium is shifted towards the products. Conversely, a small \(K\) means equilibrium favors the reactants.

• Given equilibrium constant values for individual reactions, these constants can be manipulated and combined to find the \(K\) for the desired overall reaction.

The key to solving such problems is understanding how equilibrium constants relate to the stoichiometry of their reactions and how they change upon equation manipulation. It's an important underlying principle of chemical kinetics and thermodynamics, bridging our understanding between dynamic chemical processes and their stable states.

Thermodynamics

Thermodynamics in chemistry is the study of energy changes, especially heat exchange, in chemical reactions and physical changes of state. It provides insights into whether a process will occur spontaneously.

• A spontaneous reaction, under certain conditions, will proceed without external input due to favorable energy changes, often related to equilibrium.

In the context of chemical equilibrium, thermodynamics helps us understand how changes in temperature, pressure, or concentration will shift an equilibrium. According to Le Chatelier's principle:

• If a system at equilibrium is disturbed, it will adjust in a way to counteract the disturbance and restore equilibrium.

Equilibrium constants are intrinsically linked to the Gibbs free energy change (\(\Delta G\)) of a reaction.Using the relationship:

• \(\Delta G = -RT \ln K\), where \(R\) is the gas constant and \(T\) is temperature, we can determine the spontaneity of the reaction.

A negative \(\Delta G\) means a reaction will proceed spontaneously, aligning with a \(K\) greater than 1. Understanding thermodynamics enables us to predict the behavior of reactions under different conditions and explains the fundamental principles governing the equilibrium constants.