A sample of nitrosyl bromide (NOBr) decomposes according to the equation $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ An equilibrium mixture in a \(5.00-\mathrm{L}\) vessel at \(100{ }^{\circ} \mathrm{C}\) contains \(3.22 \mathrm{~g}\) of \(\mathrm{NOBr}, 3.08 \mathrm{~g}\) of \(\mathrm{NO},\) and \(4.19 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) (a) Calculate \(K_{c}\) (b) What is the total pressure exerted by the mixture of gases? (c) What was the mass of the original sample of NOBr?

To calculate the moles of each species at equilibrium, we will use their given masses and molar masses: Molar mass of NOBr = 14.0067 g/mol (N) + 15.999 g/mol (O) + 79.904 g/mol (Br) = 109.910 g/mol Moles of NOBr = 3.22 g / 109.910 g/mol = 0.0293 mol Molar mass of NO = 14.0067 g/mol (N) + 15.999 g/mol (O) = 30.006 g/mol Moles of NO = 3.08 g / 30.006 g/mol = 0.1027 mol Molar mass of Br2 = 2 * 79.904 g/mol (Br) = 159.808 g/mol Moles of Br2 = 4.19 g / 159.808 g/mol = 0.0262 mol

To obtain the equilibrium concentrations for each species, we will divide their moles by the volume of the container (5.00 L): [NOBr] = 0.0293 mol / 5.00 L = 0.00586 M [NO] = 0.1027 mol / 5.00 L = 0.02054 M [Br2] = 0.0262 mol / 5.00 L = 0.00524 M

Using the balanced chemical equation and the concentrations at equilibrium, we can calculate the equilibrium constant Kc: Kc = [NO]^2[Br2] / [NOBr]^2 = (0.02054)^2 * (0.00524) / (0.00586)^2 = 0.157

The total moles of gas in the container at equilibrium are: Total moles = Moles of NOBr + Moles of NO + Moles of Br2 = 0.0293 + 0.1027 + 0.0262 = 0.1582 mol Now, we can use the ideal gas law to find the total pressure: PV = nRT Total pressure P = nRT / V = (0.1582 mol) * (0.0821 L * atm / mol * K) * (373 K) / (5.00 L) = 1.021 atm

The balanced chemical equation tells us that, for each mole of NO formed, one mole of NOBr decomposes. Therefore, the number of moles of NOBr that decomposed is equal to the number of moles of NO at equilibrium: Moles of NOBr decomposed = Moles of NO at equilibrium = 0.1027 mol So the original number of moles of NOBr was: Original moles of NOBr = Moles of NOBr at equilibrium + Moles of NOBr decomposed = 0.0293 mol + 0.1027 mol = 0.1320 mol Now we can convert this back to mass: Mass of the original sample of NOBr = 0.1320 mol * 109.910 g/mol = 14.51 g #Conclusion# The equilibrium constant Kc is 0.157, the total pressure exerted by the equilibrium mixture is 1.021 atm, and the mass of the original sample of nitrosyl bromide was 14.51 g.