Iodine chloride decomposes at high temperatures to iodine and chlorine gases. $$2 \mathrm{ICl}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Cl}_{2}(g)$$ Equilibrium is established at a certain temperature when the partial pressures of ICI, \(\mathrm{I}_{2}\), and \(\mathrm{Cl}_{2}\) are (in atm) \(0.43,0.16\), and \(0.27\), respectively. (a) Calculate \(K\). (b) If enough iodine condenses to decrease its partial pressure to \(0.10 \mathrm{~atm}\), in which direction will the reaction proceed? What is the partial pressure of iodine when equilibrium is re-established?

First, we need to write the equilibrium constant expression for the decomposition reaction of iodine chloride: $$K_p = \frac{[\mathrm{I}_{2}][\mathrm{Cl}_{2}]}{[\mathrm{ICl}]^2}$$ Notice that we're using \(K_p\) since the question provides the partial pressures.

Next, plug in the given partial pressures of ICl (\(0.43\,\mathrm{atm}\)), I2 (\(0.16\,\mathrm{atm}\)), and Cl2 (\(0.27\,\mathrm{atm}\)) into the equilibrium expression and calculate K: $$K_p = \frac{(0.16)(0.27)}{(0.43)^2} = 0.261$$

Now, if enough iodine condenses and its partial pressure is decreased to \(0.10\,\mathrm{atm}\), we can calculate the reaction quotient (Q) using the new partial pressure of I2: $$Q = \frac{(0.10)(0.27)}{(0.43)^2} = 0.162$$

Compare the computed Q value to the given K value to determine in which direction the reaction will proceed. Since \(Q < K\), the reaction will shift towards the products (I2 and Cl2), meaning that more iodine chloride will decompose.

Let \(x\) be the amount of iodine formed, then the same amount of chlorine will also be formed while twice the amount of iodine chloride will decompose. We can express the new equilibrium in terms of \(x\) using the new partial pressures: Partial pressure of ICl: \(0.43 - 2x\) Partial pressure of I2: \(0.10 + x\) Partial pressure of Cl2: \(0.27 + x\) Now, use the equilibrium constant K and the new partial pressures to solve for \(x\): $$0.261 = \frac{[(0.10 + x)(0.27 + x)]}{(0.43 - 2x)^2}$$ Solving this quadratic equation, we get two possible values for \(x\): \(0.0234\) or a negative value (which we can disregard as partial pressures cannot be negative). So, the correct value of \(x\) is \(0.0234\,\mathrm{atm}\). Finally, calculate the new partial pressure of iodine when equilibrium is re-established: New partial pressure of I2: \(0.10 + 0.0234 = 0.1234\,\mathrm{atm}\)