At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$ If 2.0 moles of NO and 1.0 mole of \(\mathrm{Cl}_{2}\) are placed into a \(1.0-\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.

For the given reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$ The equilibrium expression, K, is: $$K = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}$$ Where K is given as \(1.6 \times 10^{-5}\) at \(35^{\circ}\mathrm{C}\).

We are given the initial amounts (in moles) of NO and Cl₂ and the fixed volume of the flask. Therefore, we can calculate the initial concentrations: $$[\mathrm{NO}]_{0} = \frac{2.0 \, \text{moles}}{1.0 \, \text{L}} = 2.0 \, \text{M}$$ $$[\mathrm{Cl}_{2}]_{0} = \frac{1.0 \, \text{moles}}{1.0 \, \text{L}} = 1.0 \, \text{M}$$ $$[\mathrm{NOCl}]_{0} = 0 \, \text{M}$$ Now, set up the ICE table: | | NOCl | NO | Cl₂ | |------|------|----|-----| | I | 0 | 2.0| 1.0 | | C | +2x | -2x| -x | | E | 2x | 2.0-2x| 1.0-x | Notice that the stoichiometry of the reaction shows equal moles of NOCl converting into NO, so the change in concentrations will follow a ratio of 2:2, or 1:1.

Using K and the ICE table values, we have: \(K = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2} = \frac{(2.0-2x)^2(1.0-x)}{(2x)^2}\) Plug the given value of K: \(1.6 \times 10^{-5} = \frac{(2.0-2x)^2(1.0-x)}{(2x)^2}\) To solve for x, quadratic equation solvers can be used. The value of x will give the changes in the concentrations of the reactants and products at equilibrium. Upon solving, we get: \(x = 0.0016\) Now, plug the value of x into the E row of the ICE table: $$[\mathrm{NOCl}] = 2x = 2(0.0016) = 0.0032 \, \text{M}$$ $$[\mathrm{NO}] = 2.0 - 2x = 2.0 - 2(0.0016) = 1.9968 \, \text{M}$$ $$[\mathrm{Cl}_{2}] = 1.0 - x = 1.0 - 0.0016 = 0.9984 \, \text{M}$$ So, the equilibrium concentrations are: $$[\mathrm{NOCl}] = 0.0032 \, \text{M}$$ $$[\mathrm{NO}] = 1.9968 \, \text{M}$$ $$[\mathrm{Cl}_{2}] = 0.9984 \, \text{M}$$