For the reaction $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$,it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: \([\mathrm{NO}(g)]=8.1 \times 10^{-3} \mathrm{M}\) \(\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} \mathrm{M},\left[\mathrm{N}_{2}(g)\right]=5.3 \times 10^{-2} \mathrm{M},\) and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=\) \(2.9 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

Short Answer

Expert verified
The value of the equilibrium constant \(K\) for the given reaction at this temperature is approximately \(4.04 \times 10^{6}\).

Step by step solution

01

Write the expression for the equilibrium constant K

Recall that the equilibrium constant, \(K\), can be expressed as follows: $$K = \frac{[\mathrm{Products}]^{\mathrm{their \, stoichiometric \, coefficients}}}{[\mathrm{Reactants}]^{\mathrm{their \, stoichiometric \, coefficients}}}$$ For the given reaction, the expression for \(K\) can be written as: $$K = \frac{[\mathrm{N}_{2}][\mathrm{H}_{2}\mathrm{O}]^2}{[\mathrm{NO}]^2[\mathrm{H}_{2}]^2}$$
02

Substitute the values of the concentrations

Now, let's plug in the given equilibrium concentrations into the expression for \(K\): $$K = \frac{\left( 5.3 \times 10^{-2} \right) \left( 2.9 \times 10^{-3} \right)^2}{\left( 8.1 \times 10^{-3} \right)^2 \left( 4.1 \times 10^{-5} \right)^2}$$
03

Calculate the value of K

Perform the calculations in the expression: $$K = \frac{(5.3 \times 10^{-2})(8.41 \times 10^{-6})}{(6.561 \times 10^{-5})(1.681 \times 10^{-9})}$$ $$K = \frac{4.4573 \times 10^{-7}}{1.1026 \times 10^{-13}}$$ $$K = 4.04 \times 10^{6}$$ At this temperature, the value of the equilibrium constant \(K\) for the given reaction is approximately \(4.04 \times 10^{6}\).

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is crucial for predicting how a chemical reaction will behave over time. It represents a state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. When this balance is achieved, the concentrations of reactants and products remain constant but not necessarily equal.

This does not mean the reaction has stopped—molecules continue to react in both directions at an equal rate. Grasping this dynamic yet stable condition is essential, and recognizing that equilibrium can be reached from either direction of the reaction is key. Environmental conditions such as temperature and pressure can shift this equilibrium, exemplified by Le Chatelier's Principle.
Reaction Quotient
The reaction quotient, denoted as Q, can seem similar to the equilibrium constant, but it plays a unique role in evaluating reaction progress. Q is calculated by the same equation used for K, utilizing the current concentrations of the reactants and products regardless of whether the system is at equilibrium.

Comparing the reaction quotient to the equilibrium constant allows us to predict which way a reaction will proceed to reach equilibrium. If Q is less than K, the forward reaction is favored to produce more products. Conversely, if Q is greater than K, the reverse reaction is favored. When Q equals K, the reaction is at equilibrium, signifying no net change in concentration over time.
Equilibrium Concentrations
Determining equilibrium concentrations is a typical endeavor in chemical kinetics and is pivotal in calculating K. These concentrations are those of the reactants and products when a reaction has reached equilibrium. In experiments or solved problems, these values are given or calculated to examine the reaction's conditions at equilibrium.

With equilibrium concentrations known, we can dissect a reaction's tendency to produce reactants or products under specific conditions. They are also invaluable for calculating reaction yields or understanding the efficiencies of different reaction pathways. When tackling problems or experimental data, interpreting these concentrations correctly can afford profound insights into the system's behavior.
Stoichiometry
The concept of stoichiometry is foundational in chemistry, encompassing the quantitative relationships between reactants and products in a chemical reaction. It bridges the gap between the molecular world and measurable quantities like mass and volume.

Stoichiometric coefficients, derived from a balanced chemical equation, dictate the proportions in which substances react and form products. They drive the calculations for determining the equilibrium constant, ensuring that we correctly account for the influence of each substance's concentration to the power of its coefficient. A clear understanding of stoichiometry is not only vital for precise calculations but also instrumental in predicting how much product can be generated from given amounts of reactants under standard or varied conditions.

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Most popular questions from this chapter

At \(327^{\circ} \mathrm{C},\) the equilibrium concentrations are \(\left[\mathrm{CH}_{3} \mathrm{OH}\right]=\) \(0.15 M,[\mathrm{CO}]=0.24 M,\) and \(\left[\mathrm{H}_{2}\right]=1.1 M\) for the reaction $$\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)$$.Calculate \(K_{\mathrm{p}}\) at this temperature.

Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$.equilibrium constants \(\mathbf{r}_{\mathbf{p}}\).\(300^{\circ} \mathrm{C}, \quad 4.34 \times 10^{-3}\) \(500^{\circ} \mathrm{C}, \quad 1.45 \times 10^{-5}\) \(600^{\circ} \mathrm{C}, \quad 2.25 \times 10^{-6}\) Is the reaction exothermic or endothermic?

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$\mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(aq)+\mathrm{H}_{2}(a q)$$. =Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10},\) what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

Consider the following statements: "Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g),\) for which at equilibrium \([\mathrm{A}]=2 M\) \([\mathrm{B}]=1 M,\) and \([\mathrm{C}]=4 \mathrm{M} .\) To a \(1-\mathrm{L}\) container of the system at equilibrium, you add 3 moles of B. A possible equilibrium condition is \([\mathrm{A}]=1 M,[\mathrm{B}]=3 M,\) and \([\mathrm{C}]=6 \mathrm{M}\) because in both cases \(K=2 . "\) Indicate everything that is correct in these statements and everything that is incorrect. Correct the incorrect statements, and explain.

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C},\) equilibrium is reached when \(50.0 \%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

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