Question

At \(165^{\circ} \mathrm{C}\), a solid chloride \((\mathrm{X})\) reacts with \(0.0417 \mathrm{~mol}\) of steam to form the solid oxychloride, \(\mathrm{Y},\) and \(\mathrm{HCl}\) gas. Steam and HCl have a 1: 2 stoichiometric ratio in the reaction $$ \mathrm{X}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Y}(s)+2 \mathrm{HCl}(g) $$ When the reaction reaches equilibrium, the gases are transferred without loss to a \(2.50-\mathrm{L}\) vessel. The temperature is kept at \(165^{\circ} \mathrm{C}\). Sufficient silver(I) ion is added to precipitate the chloride as AgCl. If \(7.29 \mathrm{~g}\) of \(\mathrm{AgCl}\) are obtained, what is \(K\) for the reaction of the chloride with steam at \(165^{\circ} \mathrm{C} ?\)

Short answer

The equilibrium constant, K, for the reaction between a solid chloride (X) and steam at \(165^{\circ} \mathrm{C}\) is 0.0640.

Step-by-step solution

Convert the mass of AgCl to moles of HCl

Using the molar mass of AgCl (107.87 g/mol for Ag and 35.45 g/mol for Cl, total 143.32 g/mol), we convert the mass of AgCl to moles and relate it to the moles of HCl produced according to the stoichiometry of the precipitation reaction, where Ag+ + Cl- -> AgCl. $$ \mathrm{moles \;of \;AgCl} = \frac{7.29 \mathrm{~g}}{143.32 \mathrm{~g/mol}} = 0.0509 \mathrm{~moles} $$ Since 1 mole of HCl gives 1 mole of AgCl during precipitation, the moles of HCl produced in the reaction are equal to the moles of AgCl. $$ \mathrm{moles \;of \;HCl} = 0.0509 \mathrm{~moles} $$

Calculate the moles of H2O and HCl at equilibrium

In the original reaction, 1 mole of steam (H2O) reacts with 2 moles of HCl. Since we started with 0.0417 moles of steam, we can write a stoichiometric equation using the reaction coefficients: $$ \mathrm{moles \;of \;H_{2} O \;at \;equilibrium} = 0.0417 - x $$ $$ \mathrm{moles \;of \;HCl \;at \;equilibrium} = 2x $$ We now know that in the equilibrium state, there are 0.0509 moles of HCl. We can use this information to calculate the amount of steam (H2O) and HCl reacted (x): $$ 2x = 0.0509 $$ $$ \Rightarrow x = 0.02545 $$ Therefore: $$ \mathrm{moles \;of \;H_{2} O \;at \;equilibrium} = 0.0417 - 0.02545 = 0.01625 $$

Write the expression for the equilibrium constant, K

The expression for the equilibrium constant, K, for the given reaction is: $$ K = \frac{[\mathrm{HCl}]^2}{[\mathrm{H_{2}O}]} $$

Calculate the value of K

Convert the moles of H2O and HCl at equilibrium into concentrations by dividing them by the volume of the vessel (2.50 L): $$ \mathrm{[H_{2}O]} = \frac{0.01625 \mathrm{~moles}}{2.50 \mathrm{~L}} = 0.00650 \mathrm{~M} $$ $$ \mathrm{[HCl]} = \frac{0.0509 \mathrm{~moles}}{2.50 \mathrm{~L}} = 0.02036 \mathrm{~M} $$ Now, substitute these concentrations into the expression for K and calculate its value: $$ K = \frac{(0.02036)^2}{0.00650} = 0.0640 $$ So, the equilibrium constant K for the reaction of the chloride with steam at \(165^{\circ} \mathrm{C}\) is 0.0640.