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.

First, we'll write the balanced reaction and set up an ICE table to determine the amounts of reactants and products before and after equilibrium is reached. Reaction: \(2\thinspace NOBr(g) \rightleftharpoons 2\thinspace NO(g) + Br_2(g)\) ICE Table: | | Initial | Change | Equilibrium | |---------|---------|----------|--------------| | NOBr | X | -2Y | X - 2Y | | NO | 0 | +2Y | 2Y | | Br2 | 0 | +Y | Y | *X represents the initial moles of NOBr, Y represents the moles of NO and Br2 at equilibrium.

We are given the initial density of the system (\(4.495\thinspace g/L\)) and the density at equilibrium (\(4.086\thinspace g/L\)). To relate the densities to the ICE table, we need to convert them to concentrations. First, we'll find the molar mass of NOBr: Molar mass of NOBr = Atomic mass of N + Atomic mass of O + Atomic mass of Br = 14.01 + 16.00 + 79.90 = \(109.91\thinspace g/mol\) Now, let's convert the initial and equilibrium densities to molar concentrations: Initial concentration of NOBr: \[\frac{4.495\thinspace g/L}{109.91\thinspace g/mol} = 0.0409\thinspace M\] Equilibrium concentration of NOBr: \[\frac{4.086\thinspace g/L}{109.91\thinspace g/mol} = 0.0371\thinspace M\] Now we can replace the initial concentration of NOBr in the ICE table: | | Initial | Change | Equilibrium | |---------|---------|----------|--------------| | NOBr | 0.0409 | -2Y | 0.0409 - 2Y | | NO | 0 | +2Y | 2Y | | Br2 | 0 | +Y | Y |

To calculate the equilibrium constant \(K\), we can use the equilibrium concentrations found in the ICE table: \[K = \frac{[NO]^2[Br_2]}{[NOBr]^2}\] Notice the change in moles for the initial and equilibrium concentrations of NOBr: 0.0409 - 0.0371 = 0.0038 M Now, we can substitute the values of Y: 2Y = 0.0038/2 = 0.0019 Y = 0.0019 The equilibrium concentrations for each component in the reaction are: [NOBr] = 0.0409 - 2(0.0019) = 0.0371 M [NO] = 2(0.0019) = 0.0038 M [Br2] = 0.0019 M Now, we can plug these values into the equation for K: \[K = \frac{[NO]^2[Br_2]}{[NOBr]^2} = \frac{(0.0038)^2(0.0019)}{(0.0371)^2} = 0.0133\]

When Argon (Ar) is added to the system at equilibrium, it doesn't participate in the reaction as it's a noble gas. Therefore, it won't change the concentration of the reactants or products. Consequently, the equilibrium position will remain unchanged. As a result, the addition of \(\operatorname{Ar}(g)\) doesn't affect the equilibrium constant \(K\) value. In summary, the addition of \(\operatorname{Ar}(g)\) to the system at equilibrium does not affect the equilibrium position or the value of the equilibrium constant, \(K\).