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 \(\mathrm{Ar}(g)\) 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.

Step by step solution

01

Calculate initial moles of NOBr

First, we need to find the initial moles of NOBr. To do this, we'll use the initial density and the molar mass of NOBr. The molar mass of NOBr is: Molar mass of NOBr = \(30.01 \ \mathrm{g/mol} + 16.00 \ \mathrm{g/mol} + 79.90 \ \mathrm{g/mol} = 125.91 \ \mathrm{g/mol}\) Using the initial density, we can determine the initial moles of NOBr per liter: Initial moles of NOBr/L = \( \frac{4.495 \ \mathrm{g/L}}{125.91 \ \mathrm{g/mol}} = 0.0357 \ \mathrm{mol/L}\)

02

Calculate equilibrium moles of NOBr

Next, we'll find the equilibrium moles of NOBr. This can be done using the equilibrium density and the molar mass of NOBr: Equilibrium moles of NOBr/L = \( \frac{4.086 \ \mathrm{g/L}}{125.91 \ \mathrm{g/mol}} = 0.0324 \ \mathrm{mol/L}\)

03

Calculate moles of NO and Br2 at equilibrium

Since 2 moles of NOBr decompose to produce 2 moles of NO and 1 mole of Br2, we can use the change in NOBr concentration to determine the equilibrium concentrations of NO and Br2: \( \Delta \) NOBr = Initial moles of NOBr/L - Equilibrium moles of NOBr/L = 0.0357 - 0.0324 = 0.0033 mol/L Equilibrium moles of NO/L = 2 × \( \Delta \) NOBr = 2 × 0.0033 = 0.0066 mol/L Equilibrium moles of Br2/L = 1 × \( \Delta \) NOBr = 1 × 0.0033 = 0.0033 mol/L

04

Calculate the equilibrium constant K

Now we can use the equilibrium concentrations to calculate the equilibrium constant K for the reaction: \( K = \frac{[\mathrm{NO}]^2[\mathrm{Br_2}]}{[\mathrm{NOBr}]^2} = \frac{(0.0066)^2\cdot (0.0033)}{(0.0324)^2} = 0.0145\) So, the equilibrium constant K for the reaction is 0.0145.

05

Analyzing the effect of adding Ar(g) on the equilibrium position and the value of K

When Ar(g) is added to the system, the total pressure of the system will increase but the partial pressures of the other gases (NOBr, NO, and Br2) will not change. This is because Ar(g) is inert and does not participate in the reaction. As a result, the equilibrium position will not shift, and the value of K will remain constant. In conclusion, the equilibrium constant K for the given reaction is 0.0145. When Ar(g) is added to the system, the equilibrium position does not shift and the value of K remains the same, as the partial pressures of the other gases do not change.

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