Question

Consider the dissociation of iodine: $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ A \(1.00-\mathrm{g}\) sample of \(\mathrm{I}_{2}\) is heated at \(1200^{\circ} \mathrm{C}\) in a 500 -mL flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in problem \(15.65(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.

Short answer

The equilibrium constant \(KP\) for the reaction is approximately 199.27.

Step-by-step solution

Analyze the Provided Information

From the equation, it can be seen that for each mole of \(\mathrm{I}_{2}\) that dissociates, 2 moles of \(\mathrm{I}\) are formed. Also, it's noted that a \(1.00-\mathrm{g}\) sample of \(\mathrm{I}_{2}\) is placed in a 500 mL flask and heated to achieve a total equilibrium pressure of 1.51 atm.

Calculate the Initial Moles of \(\mathrm{I}_{2}\)

First, calculate the initial moles of \(\mathrm{I}_{2}\) by using the molar mass. So, the initial moles of \(\mathrm{I}_{2}\) is \(1.00 g / 253.8 g/mol = 0.00394 mol\). Since the volume of the flask is 500 mL or 0.5 L, the initial concentration of \(\mathrm{I}_{2}(g)\) is then given by \(0.00394 mol / 0.5 L = 0.00788 M\).

Calculate the Degree of Dissociation (\(\alpha\))

The degree of dissociation can be calculated using the formula \(\alpha = 1-(p_{I_2}/P_t)\), where \(p_{I_2}\) is the partial pressure of \(\mathrm{I}_{2}\) and \(P_t\) is the total pressure at equilibrium. Since we calculated the partial pressure expecting none dissociation to be \(0.00788 atm\) and the total pressure is given as \(1.51 atm\), we can plug in these values to find that \(\alpha = 1 - (0.00788 atm / 1.51 atm) = 0.9948.\)

Calculate the Equilibrium Constant (KP)

The equilibrium constant for the reaction can be calculated using the formula for KP which is \(KP = \alpha^2/(1-\alpha)\). Substituting \(\alpha=0.9948\) into formula gives \(KP=0.9948^2/(1-0.9948)\approx 199.27\).

Key concepts

Iodine Dissociation

In chemical reactions, **dissociation** refers to the process where molecules break apart into smaller units, often atoms or ions. For iodine, the reaction can be represented as \[ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) \]In this reaction, iodine gas (\( \mathrm{I}_2 \)) dissociates into individual iodine atoms (\( \mathrm{I} \)). This breaking of bonds requires energy and occurs under specific conditions, like higher temperatures. Understanding this process is vital for various applications, including chemical manufacturing and analysis.
The dissociation of compounds like iodine showcases concepts like reaction dynamics and helps in calculating related parameters such as equilibrium constants.

Equilibrium Pressure

**Equilibrium pressure** in a chemical reaction is the total pressure exerted by the gaseous components in a closed system when the reaction has reached equilibrium. This means that the rates of the forward and reverse reactions are equal, resulting in stable concentrations of reactants and products.
When iodine dissociates at equilibrium, both the iodine molecules (\( \mathrm{I}_2 \)) and iodine atoms (\( \mathrm{I} \)) contribute to this pressure.

• The total pressure can be measured using a pressure gauge.
• Knowing the total pressure allows us to calculate the partial pressures of individual components.

These pressures help in determining other key properties of the reaction, such as the degree of dissociation.

Degree of Dissociation

The **degree of dissociation** (\( \alpha \)) is a measure of how much a substance has dissociated in comparison to the entire amount present. It's defined by the formula:\[ \alpha = 1 - \left( \frac{p_{\mathrm{I}_2}}{P_t} \right) \]where \( p_{\mathrm{I}_2} \) is the partial pressure of the original iodine gas, and \( P_t \) is the total equilibrium pressure.
A high degree of dissociation means most of the original iodine molecules have converted into atoms, whereas a low value indicates limited dissociation.

• It provides insight into the extent of the reaction.
• It helps calculate equilibrium constants, representing how dynamic the reaction is.

Equilibrium Constant

The **equilibrium constant** (\( K_P \)) quantifies the ratio of products to reactants at equilibrium for a gas-phase reaction. It provides insight into the favorability and extent of the reaction. For the iodine dissociation, we use:\[K_P = \frac{\alpha^2}{1-\alpha}\]Plugging the degree of dissociation (\( \alpha \)) into the formula, we find the value of \( K_P \), which indicates the position of equilibrium and whether the formation of products is favored.
This constant is crucial:

• For predicting reaction behavior under different conditions.
• For understanding how changes in conditions, like temperature or pressure, can affect the equilibrium state.

Knowing \( K_P \) helps chemists design processes efficiently by targeting desired equilibrium states.