Question

Hydrogen iodide gas decomposes to hydrogen gas and iodine gas: $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ To determine the equilibrium constant of the system, identical one-liter glass bulbs are filled with \(3.20 \mathrm{~g}\) of HI and maintained at a certain temperature. Each bulb is periodically opened and analyzed for iodine formation by titration with sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) $$ \mathrm{I}_{2}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}{ }^{2-}(a q)+2 \mathrm{I}^{-}(a q) $$ It is determined that when equilibrium is reached, \(37.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) is required to titrate the iodine. What is \(K\) at the temperature of the experiment?

Short answer

Answer: The equilibrium constant (K) for the decomposition reaction of hydrogen iodide (HI) into hydrogen gas (H2) and iodine gas (I2) under the given conditions is 0.018.

Step-by-step solution

Calculate the initial number of moles of HI

We are given that 3.20 g of HI is used in the experiment, and we know the molar mass of HI is 127.9 g/mol (1 mol of H = 1g/mol, and 1 mol of I = 126.9 g/mol). To find the initial moles of HI, use the formula: moles = mass / molar mass moles of HI = \(\frac{3.20 \, g}{127.9 \, g/\mathrm{mol}} = 0.0250 \, \mathrm{mol}\)

Determine moles of I2 formed at equilibrium

From the titration data, we know that 37.0 mL of 0.200 M Na2S2O3 is used. To find the moles of I2 at equilibrium, we use the stoichiometry of the titration reaction: $$ \mathrm{I}_{2}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}{ }^{2-}(a q)+2 \mathrm{I}^{-}(a q) $$ From the stoichiometry, 1 mol of I2 reacts with 2 moles of S2O3^(2-). So, moles of I2 = moles of S2O3^(2-) / 2 moles of S2O3^(2-) = 0.200 mol/L × 0.0370 L = 0.00740 mol moles of I2 = 0.00740 mol / 2 = 0.00370 mol

Calculate moles of HI and H2 at equilibrium

According to the reaction stoichiometry, for every mole of I2 formed, 1 mole of H2 is formed, and 2 moles of HI are consumed. Thus, we can set up the following relationships: moles of H2 = moles of I2 = 0.00370 mol moles of HI consumed = 2 × moles of I2 = 2 × 0.00370 mol = 0.00740 mol moles of HI at equilibrium = initial moles of HI - moles of HI consumed = 0.0250 mol - 0.00740 mol = 0.0176 mol

Calculate equilibrium concentrations of the involved species

Since the reaction happens in a 1-liter glass bulb, the equilibrium concentrations can be calculated as: [HI] = moles of HI / volume = 0.0176 mol / 1 L = 0.0176 M [H2] = moles of H2 / volume = 0.00370 mol / 1 L = 0.00370 M [I2] = moles of I2 / volume = 0.00370 mol / 1 L = 0.00370 M

Calculate the equilibrium constant (K)

Using the equilibrium concentrations, we can now calculate the equilibrium constant (K) for the given reaction: $$ K = \frac{[\mathrm{H}_{2}][\mathrm{I}_{2}]}{[\mathrm{HI}]^{2}} $$ K = \(\frac{(0.00370 \, \mathrm{M})(0.00370 \, \mathrm{M})}{(0.0176 \, \mathrm{M})^{2}} = 0.018\) The equilibrium constant, K, for the given decomposition reaction at the specific temperature is 0.018.

Key concepts

Stoichiometry

In chemistry, stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. It is the foundation for understanding and predicting the amounts of substances consumed and produced in a given reaction. To visualize stoichiometry, think of it like a recipe, where exact amounts of ingredients (reactants) are combined to make a fixed number of cakes (products).

In the exercise we're discussing, stoichiometry plays a crucial role. Here, it helps us determine the amount of iodine (I₂) produced from the decomposition of hydrogen iodide (HI). Knowing the original amount of HI and the stoichiometric ratios of the reactants and products, we can calculate how much iodine will form which, in turn, helps in finding the equilibrium constant, K. Remember, the ratios are dictated by the balanced chemical equation, ensuring that matter is conserved.

Chemical Equilibrium

Chemical equilibrium is the state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. As a result, the concentration of reactants and products remain constant over time, not because the reactions have stopped, but because they proceed at equal and opposite rates.

In the hydrogen iodide decomposition mentioned earlier, the forward and reverse reactions eventually reach a balance at equilibrium, which is described by the equilibrium constant, K. This important value indicates the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients. The equilibrium constant gives us insight into the position of the equilibrium: a high K value means a reaction that favors products, while a low K implies a reaction that favors reactants. The calculation of K is an essential skill when studying chemical reactions, as it aids in predicting how the reaction will respond to changes in conditions.

Titration Analysis

Titration is an analytical technique used to determine the concentration of a solute in a solution. It involves a reaction between a solution of known concentration (the titrant) and the solution of the unknown concentration (the analyte) to achieve a neutralization or complete reaction.

In the equilibrium exercise, titration analysis is employed to find out the amount of iodine formed at equilibrium. Sodium thiosulfate solution is used as the titrant to react with the iodine. The stoichiometry of the titration reaction tells us that it takes two moles of sodium thiosulfate to react with one mole of iodine. By measuring precisely how much titrant is needed to react completely with the iodine in the sample, we can back-calculate to find the concentration of iodine. This data is critical as it is used to ascertain the equilibrium concentrations necessary for calculating the equilibrium constant, K, for the reaction.