In a given experiment, 5.2 moles of pure NOCl was placed in an otherwise empty \(2.0-\mathrm{L}\) container. Equilibrium was established by the following reaction: $$2 \operatorname{Nocl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K=1.6 \times 10^{-5}$$, a. Using numerical values for the concentrations in the Initial row and expressions containing the variable \(x\) in both the Change and Equilibrium rows, complete the following table summarizing what happens as this reaction reaches equilibrium. Let \(x=\) the concentration of \(\mathrm{Cl}_{2}\) that s present at equilibrium. b. Calculate the equilibrium concentrations for all species.

First, let's convert the initial moles of NOCl into concentration by dividing it by the volume of the container : Initial concentration of NOCl = \(\frac{5.2 \, moles}{2.0 \, L}\) = \(2.6 \, M\) Now, we will construct an ICE table using the given information: | Species | Initial | Change | Equilibrium | |:-----------:|:-------:|:-------:|:-----------:| | NOCl | 2.6 M | -2x M | 2.6 - 2x M | | NO | 0 M | 2x M | 2x M | | Cl₂ | 0 M | x M | x M |

Next, we will write the expression for the equilibrium constant, K, using the concentrations of species from the ICE table: \(K = \frac{[NO]^2[Cl_{2}]}{[NOCl]^2}\) Plug in the numerical value of K and the concentrations from the ICE table: \(1.6 \times 10^{-5} = \frac{(2x)^2(x)}{(2.6-2x)^2}\)

Now, we need to solve this equation for x. This will give us the concentration of Cl₂ at equilibrium: First, simplify the equation: \(1.6 \times 10^{-5} = \frac{4x^3}{(2.6-2x)^2}\) To make it easier to solve, we can cross-multiply: \(1.6 \times 10^{-5}(2.6-2x)^2 = 4x^3\) This equation is a cubic equation in x. Solving this equation can be quite complex, so we can make an approximation by assuming that the value of x will be much smaller than 2.6, thus: \(1.6 \times 10^{-5} \approx \frac{4x^3}{(2.6)^2}\) Solve for x: \(x^3 = \frac{(1.6\times 10^{-5})(2.6)^2}{4}\) \(x^3 = \frac{1.08704 \times 10^{-4}}{4}\) \(x^3 = 2.7176 \times 10^{-5}\) Thus, \(x = \sqrt[3]{2.7176 \times 10^{-5}}\) \(x \approx 3.0 \times 10^{-2}\, M\)

Now, we will use the value of x to calculate the equilibrium concentrations of all species in the reaction: Equilibrium concentration of NOCl: \[2.6 - 2x\, M = 2.6 - 2(3.0 \times 10^{-2})\, M= 2.6 - 0.06\, M = 2.54\, M\] Equilibrium concentration of NO: \[2x\, M = 2(3.0 \times 10^{-2})\, M = 0.06\, M \] Equilibrium concentration of Cl₂: \[x\, M = 3.0 \times 10^{-2}\, M\]

The equilibrium concentrations of all species are as follows: NOCl: 2.54 M NO: 0.06 M Cl₂: 3.0 x 10⁻² M