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-L vessel at \(100^{\circ} \mathrm{C}\) contains \(3.22 \mathrm{~g}\) of 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?

First, let's determine the number of moles for NOBr, NO, and Br₂ by using their respective molar masses. Molar masses: NOBr: 14.01(g/mol for N) + 15.999(g/mol for O) + 79.904(g/mol for Br) = 109.923 g/mol NO: 14.01(g/mol for N) + 15.999(g/mol for O) = 30.009 g/mol Br₂: 2 × 79.904(g/mol for Br) = 159.808 g/mol Moles of each component: n_NOBr = 3.22 g / 109.923 g/mol = 0.0293 mol n_NO = 3.08 g / 30.009 g/mol = 0.1027 mol n_Br₂ = 4.19 g / 159.808 g/mol = 0.0262 mol

Since the volume of the reaction is given as 5.00 L, we can determine the concentrations of each component at equilibrium by dividing the number of moles by the volume. [NOBr] = n_NOBr / 5.00 L = 0.0293 mol / 5.00 L = 0.00586 M [NO] = n_NO / 5.00 L = 0.1027 mol / 5.00 L = 0.02054 M [Br₂] = n_Br₂ / 5.00 L = 0.0262 mol / 5.00 L = 0.00524 M

Now that we have the concentrations at equilibrium, we can find the equilibrium constant using the balanced chemical reaction: 2 NOBr(g) ⇌ 2 NO(g) + Br₂(g) $$K_{c} = \frac{[\mathrm{NO}]^{2}[\mathrm{Br}_{2}]}{[\mathrm{NOBr}]^{2}}$$ Plugging in the concentration values: $$K_{c} = \frac{(0.02054)^{2}(0.00524)}{(0.00586)^{2}}$$ Kc = 12.3 So, the equilibrium constant Kc = 12.3.

To find the total pressure exerted by the mixture, we can use the ideal gas law, which is given by: PV = nRT The values of R (gas constant) and T (temperature) are given as R = 0.08206 L atm / mol K and T = 100 °C = 373 K. Let's start by finding the total number of moles of gas in the mixture: n_total = n_NOBr + n_NO + n_Br₂ = 0.0293 mol + 0.1027 mol + 0.0262 mol = 0.1582 mol Now we can substitute the values into the ideal gas law equation. P * 5.00 L = 0.1582 mol * 0.08206 L atm / mol K * 373 K Solving this equation, we get: P = 1.091 atm Thus, the total pressure exerted by the mixture is 1.091 atm.

To find the original mass of NOBr, we need to determine the number of moles of NOBr that dissociated during the reaction. Since the balanced equation shows that for every 2 moles of NOBr that decompose, 2 moles of NO are formed and 1 mole of Br₂ is formed, we can compare the number of moles to determine the initial number of moles of NOBr. From the balanced equation, the decomposition ratio of NOBr to NO is 1:1. So, the change in the number of moles of NO will be equal to the change in the number of moles of NOBr decomposed. Δn_NOBr = Δn_NO = n_NO - n_NO (initial) Since n_NO (initial) = 0, we have: Δn_NOBr = 0.1027 mol Now, let's find the initial number of moles of NOBr: n_NOBr (initial) = n_NOBr + Δn_NOBr = 0.0293 mol + 0.1027 mol = 0.132 mol Finally, let's convert the moles into grams using the molar mass of NOBr: Initial mass of NOBr = n_NOBr (initial) * molar mass of NOBr = 0.132 mol * 109.923 g/mol = 14.51 g Thus, the initial mass of the nitrosyl bromide sample was 14.51 g.