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.

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, \[n_{NOBr}^{initial} = \frac{4.495\,\mathrm{g/L} \cdot V}{110.00\,\mathrm{g/mol}} = \frac{0.04086\,\mathrm{mol/L} \cdot V}{1\,\mathrm{L}} = 0.04086\,\mathrm{mol}.\]

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).

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: \[V' = \frac{0.04086\,\mathrm{mol} \cdot 110.00\,\mathrm{g/mol}}{4.086\,\mathrm{g/L}}.\]

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: \([NOBr]_{eq} = \frac{0.04086-x}{V'}\,\mathrm{M},\) \([NO]_{eq} = \frac{x}{V'}\,\mathrm{M},\) \([Br2]_{eq} = \frac{0.5x}{V'}\,\mathrm{M}.\)

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.

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.