If \(Q=1.0\) for the reaction \(N_{2}(g)+O_{2}(g) \rightarrow\) \(2 \mathrm{NO}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\), will the reaction have a tendency to form products or reactants, or will it be at equilibrium?

Short Answer

Expert verified
Without the value of K, we cannot definitively determine whether the reaction will favor the formation of products, reactants, or be at equilibrium. The tendency depends on the equilibrium constant, K, for this reaction at 25 degrees Celsius.

Step by step solution

01

Understanding the Reaction Quotient (Q)

The reaction quotient Q reflects the ratio of product concentrations to reactant concentrations at a given moment. When Q=1.0 for the reaction in question, this means that the ratio of products to reactants is currently equal to the equilibrium constant (K) for this reaction at 25 degrees Celsius.
02

Comparing Q to the Equilibrium Constant (K)

Evaluate the reaction at 25 degrees Celsius to determine K, the equilibrium constant. If Q equals K, the reaction is at equilibrium. If Q is greater than K, the reaction will proceed in the direction of the reactants to reach equilibrium. If Q is less than K, the reaction will proceed in the direction of the products to reach equilibrium.
03

Determine the Reaction's Tendency

Since we don't have a value for K, we can't directly compare Q and K. However, the question provides Q and asks about the tendency of the reaction. Without additional information, we cannot definitively determine the direction in which the reaction will proceed. It will be at equilibrium, favor the formation of products, or favor the formation of reactants depending on the value of K relative to Q.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to the state of a chemical reaction in which the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of the reactants and products remain constant over time, but not necessarily equal.

For a reaction such as
\[N_{2}(g) + O_{2}(g) \rightarrow 2NO(g),\]
at equilibrium, the process of nitrogen and oxygen combining to form nitrogen monoxide occurs at the same rate as the decomposition of nitrogen monoxide back into nitrogen and oxygen gases.

Understanding equilibrium is crucial for chemists because it tells them how different conditions, such as temperature and pressure, can affect the outcome of a reaction. In applications like industrial synthesis, achieving the right equilibrium can mean more efficient production and less wasted resources.
Equilibrium Constant
The equilibrium constant (\(K\)) is a number that quantifies the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium, each raised to the power of their stoichiometric coefficients in the balanced equation. For the given reaction:
\[N_{2}(g) + O_{2}(g) \rightarrow 2NO(g),\]
the equilibrium constant would be expressed as:\[K = \frac{[NO]^{2}}{[N_{2}][O_{2}]}.\]

The value of \(K\) indicates the extent to which a reaction proceeds. A high \(K\) value suggests that, at equilibrium, the reaction mixture is rich in products, while a low \(K\) suggests a mixture rich in reactants. It's worth noting that \(K\) is temperature-dependent and only constant at a given temperature.
Equilibrium in Chemical Reactions
Equilibrium in chemical reactions is not a static state, but rather a dynamic one where reactions continue to occur yet produce no net change in the concentrations of reactants and products. This is because the speed at which the reactants are forming products is perfectly balanced by the speed at which the products are decomposing back into reactants. This balance can be disturbed by changing conditions such as temperature, pressure, or concentration, leading to a shift in the equilibrium position.

In the exercise, the reaction quotient (\(Q\)) helps predict the direction in which the reaction will proceed in order to reach equilibrium. It is a snapshot of the reaction's status at a particular moment, and comparing \(Q\) to \(K\) provides insight into which way the reaction will move to reach equilibrium under the given conditions.
Le Chatelier's Principle
Le Chatelier's principle is a fundamental concept that helps predict how a change in conditions will affect the position of equilibrium in a chemical reaction. It states that if an external change is applied to a reaction mixture at equilibrium, the system will adjust itself to counteract that change and establish a new equilibrium.

For instance, if the pressure is increased, the equilibrium will shift to reduce the pressure, which often means favoring the direction that has fewer moles of gas. If temperature increases, the system favors the endothermic reaction to absorb the excess heat. Applying this principle allows chemists to manipulate the conditions to maximize the yield of desired products.

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Most popular questions from this chapter

At \(2500 \mathrm{~K}, K_{e}=20\) for the reaction \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}(\mathrm{g}) .\) An analysis of a reaction vessel at \(2500 \mathrm{~K}\) revealed the presence of \(0.18 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{Cl}_{2}, 0.31 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~F}_{2}\), and \(0.92 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{ClF}\). Will \(\mathrm{ClF}\) tend to form or to decompose as the reaction proceeds toward cquilibrium?

Let \(\alpha\) be the fraction of \(\mathrm{PCl}_{5}\) molecules that have decomposed to \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) in the reaction \(\mathrm{PCl}_{5}(\mathrm{~g})=\) \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) in a constant- volume container, so that the amount of \(\mathrm{PCl}_{5}\) at equilibrium is \(n(1-\alpha)\), where \(n\) is the pressure present initially. Derive an equation for \(K\) in terms of \(\alpha\) and the total pressure \(P\), and solve it for \(\alpha\) in terms of \(P\). Calculate the fraction decomposed at \(556 \mathrm{~K}\), at which temperature \(K=4.96\), and the total pressure is (a) \(0.50\) har; (b) \(1.00\) bar.

A mixture of \(0.0560 \mathrm{~mol} \mathrm{O}_{2}\) and \(0.0200 \mathrm{~mol}\) \(\mathrm{N}_{2} \mathrm{O}\) is placed in a \(1.00-\mathrm{L}\) reaction vessel at \(25^{\circ} \mathrm{C}\). When the reaction \(2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{~g})=4 \mathrm{NO}_{2}(\mathrm{~g})\) is at equilibrium, \(0.0200\) mol \(\mathrm{NO}_{2}\) is present. (a) What are the equilibrium concentrations? (b) What is the value of \(K_{c}\) ?

Write the reaction quotients \(Q_{c}\) for (a) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CuCl}_{2}(\mathrm{~s})\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) (s) \(\rightarrow \mathrm{N}_{2} \mathrm{O}\) (g) \(+2 \mathrm{H}_{2} \mathrm{O}\) (g) (c) \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{MgO}\) (s) \(+\mathrm{CO}_{2}\) (g)

Given that \(K_{c}=61\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(500 \mathrm{~K}\), calculate whether more ammonia will tend to form when a mixture of composition \(2.23 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(1.24 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(1.12 \times 10^{-4} \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{NH}_{3}\) is present in a container at \(500 \mathrm{~K}\).

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