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Consider the chemical equilibrium described by the equation $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \leftrightharpoons 2 \mathrm{NOBr}(g) $$ Use Le Châtelier's principle to predict the effect on the equilibrium concentration of \(\operatorname{NOBr}(g)\) and of \(\mathrm{NO}(g)\) resulting from (a) an increase in the concentration of \(\mathrm{Br}_{2}(g)\) (b) a twofold decrease in the reaction volume

Short Answer

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(a) Increase \(\mathrm{NOBr}(g)\), decrease \(\mathrm{NO}(g)\). (b) Increase \(\mathrm{NOBr}(g)\), decrease \(\mathrm{NO}(g)\).

Step by step solution

01

Understanding the Chemical Equilibrium

The equilibrium reaction is given by \(2 \mathrm{NO}(g) + \mathrm{Br}_2(g) \leftrightharpoons 2 \mathrm{NOBr}(g)\). According to Le Châtelier's principle, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and establish a new equilibrium.
02

Effect of Increasing \(\mathrm{Br}_2(g)\) Concentration

When the concentration of \(\mathrm{Br}_2(g)\) is increased, the equilibrium shifts towards producing more \(2 \mathrm{NOBr}(g)\) to reduce the concentration of \(\mathrm{Br}_2(g)\). Thus, the concentration of \(\mathrm{NOBr}(g)\) will increase, while the concentration of \(\mathrm{NO}(g)\) will decrease.
03

Effect of Decreasing Reaction Volume

Decreasing the reaction volume increases the pressure of the system. To reduce the pressure, the equilibrium will shift towards the side with fewer moles of gas. The reaction \(2 \mathrm{NO}(g) + \mathrm{Br}_2(g) \leftrightharpoons 2 \mathrm{NOBr}(g)\) has 3 moles on the left and 2 moles on the right. Hence, the equilibrium shifts to the right, increasing \(\mathrm{NOBr}(g)\) and decreasing \(\mathrm{NO}(g)\).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction reaches a state where the concentrations of reactants and products do not change over time. At this point, the forward and reverse reactions occur at equal rates. The chemical equation given in the exercise represents such an equilibrium:\[ 2 \text{NO}(g) + \text{Br}_2(g) \leftrightharpoons 2 \text{NOBr}(g) \]In this system, both the nitrogen monoxide (NO) and bromine (Brelements maintain a constant concentration alongside the nitric oxide bromide (NOBr). Le Châtelier's principle helps predict how changes in physical conditions affect equilibrium. If the system undergoes any disturbance like concentration, pressure, or temperature shifts, the equilibrium will adjust to minimize that stress.
Reaction Volume
Reaction volume refers to the space in which a chemical reaction takes place. It can significantly affect the equilibrium according to Le Châtelier's principle. When the volume of a reaction mixture decreases, it leads to a rise in pressure, given the gases involved.For the reaction:\[ 2 \text{NO}(g) + \text{Br}_2(g) \leftrightharpoons 2 \text{NOBr}(g) \]Decreasing the volume increases pressure, prompting the system to shift the equilibrium towards the side with fewer gas moles to counteract this change. Here, the right side has only 2 moles of NOBr, whereas the left has 3 moles (2 NO and 1 Br\Consequently, reducing the volume drives the equilibrium towards forming more NOBr, reducing - the total pressure, - the concentration of NO and Brresulting in more efficient synthesis of NOBr.
Pressure Effects
Pressure is a crucial factor in determining how a gaseous equilibrium responds to external changes. Applying Le Châtelier's principle, pressure changes alter equilibrium in a manner aimed at balancing the shifts. In our equilibrium equation:\[ 2 \text{NO}(g) + \text{Br}_2(g) \leftrightharpoons 2 \text{NOBr}(g) \]- An increase in the system's pressure, often due to decreased volume, leads to equilibrium shifting towards fewer moles of gaseous species.- Here, we'll see an increase in the formation of NOBr, which consists of fewer moles compared to NO and Br\This shift illustrates an adaptive response to counterbalance increased pressure. Understanding these shifts and adjustments is vital for predicting how a system achieves stability when faced with external pressures. Thus, paying attention to moles and their placement on either side of the equilibrium is key in determining how such pressure changes affect the chemical balance.

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Most popular questions from this chapter

At \(1000^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.263 \mathrm{bar}^{-1}\) for the equation $$ \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{4}(g) $$ Calculate the equilibrium pressure of \(\mathrm{CH}_{4}(g)\) in bars if \(0.250\) moles of \(\mathrm{CH}_{4}(g)\) is placed in a \(4.00\) -liter container at \(1000^{\circ} \mathrm{C}\).

Sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\), decomposes to \(\mathrm{SO}_{2}(g)\) and \(\mathrm{Cl}_{2}(g)\) at \(100^{\circ} \mathrm{C}\) according to $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \leftrightharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ A \(6.175\) -gram sample of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\) is placed in an evacuated 1.00-liter container at \(100^{\circ} \mathrm{C}\), and the total pressure at equilibrium is found to be \(2.41\) bar. Calculate the partial pressures of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g), \mathrm{SO}_{2}(g)\), and \(\mathrm{Cl}_{2}(g)\) and the value of \(K_{\mathrm{p}}\)

Phosgene, \(\mathrm{COCl}_{2}(g)\), a toxic gas used in the synthesis of a variety of organic compounds, decomposes according to $$ \mathrm{COCl}_{2}(g) \leftrightharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ A sample of phosgene gas at an initial concentration of \(0.500 \mathrm{M}\) is heated at \(527^{\circ} \mathrm{C}\) in a reaction vessel. At equilibrium, the concentration of \(\mathrm{CO}(g)\) was found to be \(0.046 \mathrm{M}\). Calculate the equilibrium constant for the reaction equation at \(527^{\circ} \mathrm{C}\).

Given the equilibrium constants at \(1000 \mathrm{~K}\) for the following equations (1) \(\mathrm{CaCO}_{3}(s) \leftrightharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) \(K_{\mathrm{P}_{1}}=0.040 \mathrm{bar}\) (2) \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)\) \(K_{\mathrm{P}_{2}}=1.9 \mathrm{bar}\) determine the equilibrium constant, \(K_{\mathrm{p}}\), at \(1000 \mathrm{~K}\) for the equation (3) \(\mathrm{CaCO}_{3}(s)+\mathrm{C}(s) \leftrightharpoons \mathrm{CaO}(s)+2 \mathrm{CO}(g)\)

Consider the methanation reaction described by the chemical equation $$ \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ It was found that \(0.618\) moles of \(\mathrm{CO}(g), 0.387\) moles of \(\mathrm{H}_{2}(\mathrm{~g}), 0.387 \mathrm{moles}\) of \(\mathrm{CH}_{4}(g)\), and \(0.387\) moles of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) were present in an equilibrium mixture in a 1.00-liter container. All the water vapor was removed and the system allowed to come to equilibrium again. Calculate the concentration of all gases in the new equilibrium system. (You must solve the resulting equation by trial and error.)

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