Question

Consider the decomposition at \(25^{\circ} \mathrm{C}\) of one mole of NOBr gas into \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\) gases. At equilibrium, the concentrations of \(\mathrm{NOBr}, \mathrm{NO},\) and \(\mathrm{Br}_{2}\) gases are \(0.0162 \mathrm{M}\), \(0.0011 \mathrm{M},\) and \(0.072 \mathrm{M},\) respectively. (a) Write a balanced equation for the reaction. (b) Calculate \(K\) for the reaction at \(25^{\circ} \mathrm{C}\). (Note that the gases need to be expressed as pressure in atm.)

Short answer

Based on the decomposition of NOBr gas into NO and Br2 gases and their concentrations at equilibrium, we calculated the equilibrium constant, K. The balanced equation for this reaction is NOBr (g) ⇌ NO (g) + 1/2 Br2 (g). After determining the partial pressures for each gas, the equilibrium constant K is calculated as (0.0269 atm) * (1.763 atm)^(1/2) / (0.397 atm), which results in an approximate value of 0.0412 at 25°C.

Step-by-step solution

Write the balanced equation

The balanced equation for the decomposition of NOBr gas into NO and Br2 gases is: NOBr (g) ⇌ NO (g) + 1/2 Br2 (g) #b) Calculating the equilibrium constant K#

Determine the partial pressures of each gas

We are given the concentrations of the gases in molarity, but we need their partial pressures in atm to calculate K. We can use the Ideal Gas Law, PV = nRT, to convert the concentrations into partial pressures by solving for P, where V and n are constant for all gases: P = nC/RT, where nC is the concentration of the gas, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin (25°C + 273.15 = 298.15 K). For the NOBr gas, its concentration is 0.0162 M, so its partial pressure P_NOBr = (0.0162 M)(0.0821 L atm/mol K)(298.15 K) = 0.397 atm. For the NO gas, its concentration is 0.0011 M, so its partial pressure P_NO = (0.0011 M)(0.0821 L atm/mol K)(298.15 K) = 0.0269 atm. For the Br2 gas, its concentration is 0.072 M, so its partial pressure P_Br2 = (0.072 M)(0.0821 L atm/mol K)(298.15 K) = 1.763 atm.

Calculate the equilibrium constant K

With the partial pressures at equilibrium, we can calculate the equilibrium constant, K, using the balanced equation: K = [NO] * [Br2]^1/2 / [NOBr], where [NO], [Br2], and [NOBr] are the partial pressures of the respective gases. K = (0.0269 atm) * (1.763 atm)^(1/2) / (0.397 atm) = 0.0412. So the equilibrium constant K for the reaction at 25°C is approximately 0.0412.