Question

A sample of gaseous nitrosyl bromide (NOBr) was placed in a container fitted with a frictionless, massless piston, where it decomposed at \(25^{\circ} \mathrm{C}\) according to the following equation: $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ The initial density of the system was recorded as 4.495 \(\mathrm{g} / \mathrm{L}\) . After equilibrium was reached, the density was noted to be 4.086 \(\mathrm{g} / \mathrm{L}\) . a. Determine the value of the equilibrium constant \(K\) for the reaction. b. If \(\operatorname{Ar}(g)\) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of \(K ?\) Explain each answer.

Short answer

In summary, when the nitrosyl bromide decomposition reaction reaches equilibrium, the equilibrium constant K can be calculated from the initial and equilibrium densities and found to be \(K = \frac{0.5x^3}{(0.04086-x)^2}\) (where x pertains to the amount of species formed at equilibrium). When Ar(g) is added to the system at equilibrium at constant temperature, the equilibrium position shifts towards the reactants due to an increase in total pressure, but the value of K remains constant, as it is not affected by changes in pressure.

Step-by-step solution

Calculate the moles of NOBr initially

Calculate the initial moles of NOBr by using the initial density provided. To do this, use the molar mass of NOBr, which is \(M = 14.01 + 16.00 + 79.90 = 110.00\,\mathrm{g/mol}.\) The number of moles can be found using the formula: \(n = \frac{density \cdot volume}{molar\,mass}\). Let's call the initial volume V. Then, \[n_{NOBr}^{initial} = \frac{4.495\,\mathrm{g/L} \cdot V}{110.00\,\mathrm{g/mol}} = \frac{0.04086\,\mathrm{mol/L} \cdot V}{1\,\mathrm{L}} = 0.04086\,\mathrm{mol}.\]

Define variables for the amount of product and reactant after reaching equilibrium

Since equilibrium is reached and 1 mole of NOBr decomposes to produce 1 mole of NO and 0.5 moles of Br2, we can write: \(n_{NOBr}^{equilibrium} = (0.04086 - x)\,\mathrm{mol}\) (as it loses 1 mole for each mole of NOBr decomposed), \(n_{NO}^{equilibrium} = x\,\mathrm{mol}\) (as it gains 1 mole for each mole of NOBr decomposed), \(n_{Br2}^{equilibrium} = 0.5x\,\mathrm{mol}\) (as it gains 0.5 moles for each mole of NOBr decomposed).

Calculate the equilibrium density and volume of each species

The equilibrium density is given as 4.086 g/L. To compute the total moles of the species at equilibrium: \[density_{total} = \frac{mass_{total}}{volume}.\] Let's denote the equilibrium volume as \(V'\). Rearrange the formula and find the equilibrium volume: \[V' = \frac{mass_{total}}{density_{total}}.\] From the balanced equation, we know that for every mole of NOBr reacted, 1 mole of NO and 0.5 moles of Br2 are produced. Therefore, the mass of the species at equilibrium equals the initial mass of NOBr. The equilibrium volume can be expressed as: \[V' = \frac{0.04086\,\mathrm{mol} \cdot 110.00\,\mathrm{g/mol}}{4.086\,\mathrm{g/L}}.\]

Calculate the equilibrium concentrations of each species

Now that we have the equilibrium volume (\(V'\)), we can calculate the equilibrium concentrations by dividing the equilibrium moles (for each species from Step 2) by the equilibrium volume (\(V'\)). The equilibrium concentrations will be as follows: \([NOBr]_{eq} = \frac{0.04086-x}{V'}\,\mathrm{M},\) \([NO]_{eq} = \frac{x}{V'}\,\mathrm{M},\) \([Br2]_{eq} = \frac{0.5x}{V'}\,\mathrm{M}.\)

Write the equilibrium expression and solve for x

Now, we write the equilibrium expression for K: \[K = \frac{[NO]^2_{eq} \cdot [Br2]_{eq}}{[NOBr]^2_{eq}}\] Substitute the equilibrium concentrations we found in the previous step and simplify: \[K = \frac{\left(\frac{x}{V'}\right)^2 \cdot \left(\frac{0.5x}{V'}\right)}{\left(\frac{0.04086-x}{V'}\right)^2}.\] The \(V'\) will cancel out, and we're left with a simpler equation: \[K = \frac{0.5x^3}{(0.04086-x)^2}.\] Now we need to find the value of x, which will allow us to determine K. To do so, we'll need to recall that there's a relationship between initial and equilibrium conditions, which is given by the densities: \[Initial\,density = \frac{mass_{initial}}{V},\] \[Equilibrium\,density = \frac{mass_{equilibrium}}{V'}.\] Solve these equations by substituting the given densities to find the values of V and \(V'\) respectively, to find the relationship between them: \(\frac{V}{V'} = \frac{4.086}{4.495}.\) Since the known values are density ratios, we won't be able to find the exact value of x. Nevertheless, we can use qualitative arguments from the problem statement to predict how x will affect K. #b. If Ar(g) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of K ? Explain each answer.

Part b: Effect of adding Ar(g) on the system at equilibrium

When Ar(g) is added to the system at equilibrium and at constant temperature, it will increase the total pressure of the system, but it doesn't take part in the reaction. According to Le Chatelier's principle, if we increase the pressure in a system at equilibrium, the reaction will shift in the direction that decreases the pressure. In this case, the forward reaction produces one more gaseous product than the reverse reaction (3 moles on the product side compared to 2 moles on the reactant side). As a result, the equilibrium will shift towards the reactants, which means it will favor the reverse reaction, where there will be consumption of NO and Br2, and an increase in the concentration of NOBr. However, the value of K remains constant because the stoichiometry of the reaction doesn't change, and K is not affected by a change in pressure. It represents the equilibrium state of the reaction and will only change if the temperature changes. So, when Ar(g) is added, the equilibrium position shifts to the left (towards the reactants), but the value of K remains the same.

Key concepts

Le Chatelier's Principle

Understanding Le Chatelier's Principle is essential when predicting how a system at equilibrium will respond to external changes. When a chemical system is in equilibrium, it means that the rates of the forward and reverse reactions are equal. Adding or removing reactants, products, or changing conditions like pressure or temperature can disturb this balance.

When an external change is applied, Le Chatelier's Principle states that the equilibrium will adjust to counteract the change. For instance, if pressure is increased by the addition of a non-reactive gas like Argon ( ext{Ar}(g)), the total pressure goes up, but the equilibrium constant ( K) isn't altered because the reaction stoichiometry isn't affected.

• If the pressure was due to reactant or product gases, the equilibrium might shift towards the side with fewer gas moles to reduce pressure.
• However, since Argon isn't involved in the reaction itself, the principle predicts no shift in the equilibrium position due to pressure changes from inert gases.

Equilibrium Position

The equilibrium position of a chemical reaction is the particular set of concentrations for reactants and products that remains constant over time. It's important to know that the position of equilibrium depends on the reaction conditions, including concentrations, pressure, and temperature.

Even though the equilibrium constant ( K) remains the same under constant temperature, the equilibrium position can vary. Additions like inert gases (e.g., Argon) don't directly alter the concentrations of reactants or products but can influence the system's pressure, subsequently causing the equilibrium position to shift according to Le Chatelier’s Principle.

• In our decomposition reaction, an increase in pressure due to added Argon doesn’t change the equilibrium constant.
• The principal effect observed is that the equilibrium "position" may shift depending on the dynamic of the moles of gases involved to counterbalance the changes.

Chemical Reactions

In the context of chemical reactions, we're dealing with processes where substances, called reactants, are transformed into new substances, called products. This transformation is described by a chemical equation, which in our case involves the decomposition of nitrosyl bromide ( NOBr) into nitric oxide ( NO) and bromine ( Br_2).

The balanced equation for this particular reaction demonstrates conservation of mass and the stoichiometry, meaning the ratio in which reactants and products participate.

• This is vital for calculating how changes in concentrations affect the equilibrium.
• Understanding the reaction process helps us determine what happens when external conditions, like the introduction of Argon, affect the system.

Reaction Stoichiometry

Reaction stoichiometry is the calculation of relative quantities of reactants and products in chemical reactions. This involves using the coefficients from the balanced chemical equation to calculate the progress of reactions.

In our problem, the stoichiometric coefficients tell us that for every 2 moles of NOBr that decompose, 2 moles of NO and 1 mole of Br_2 are produced. This stoichiometric relationship is crucial when calculating the equilibrium concentrations and determining the value of the equilibrium constant ( K).

• Stoichiometry helps us establish how x (change in moles) affects the overall equilibrium conditions.
• Applying stoichiometric principles assists in making accurate predictions about shifts in equilibrium position.

Gas Laws

Gas laws describe the behavior of gases in relation to temperature, pressure, volume, and amount. They are crucial in understanding equilibrium systems involving gases.

For instance, using the ideal gas law, we relate changes in the conditions of our gas system with pressure and volume. In the context of this exercise, key gas law principles help illustrate how changes in external factors like pressure from inert gas additions can influence reaction parameters.

• While the gas laws don't change the value of K, they do affect the equilibrium position, illustrating potential shifts under varied pressure environments.
• The relationship between volume and pressure as dictated by these laws helps understand why adding Argon at constant volume doesn't alter the equilibrium constant.