Question

A student studies the equilibrium $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ at a high temperature. She finds that the total pressure at equilibrium is \(40 \%\) greater than it was originally, when only \(\mathrm{I}_{2}\) was present. What is \(K\) for this reaction at that temperature?

Short answer

Question: Calculate the equilibrium constant (K) for the reaction I2(g) ⇌ 2I(g) when the total pressure at equilibrium is 40% greater than the initial pressure of I2(g). Answer: The equilibrium constant (K) for this reaction is approximately 0.683.

Step-by-step solution

Assign variables for partial pressures

Let's denote the initial partial pressure of \(\mathrm{I}_{2}\) as \(P_{\mathrm{I}_{2},0}\). At equilibrium, some of the \(\mathrm{I}_{2}\) has dissociated into 2 I atoms. We will denote the change in the partial pressures as x: - The partial pressure of \(\mathrm{I}_{2}\) will be \(P_{\mathrm{I}_{2}} = P_{\mathrm{I}_{2},0} - x\). - The partial pressure of I atoms will be \(P_{\mathrm{I}} = 2x\). Also, we know that the total pressure at equilibrium (P) is 40% greater than the initial pressure of \(\mathrm{I}_{2}\), or \(P = 1.4P_{\mathrm{I}_{2},0}\).

Write expressions for K and equilibrium total pressure

The equilibrium constant K for the given reaction can be written in terms of partial pressures: $$K = \frac{P_{\mathrm{I}}^2}{P_{\mathrm{I}_{2}}}$$ Substitute the expressions for the partial pressures found in Step 1: $$K = \frac{(2x)^2}{(P_{\mathrm{I}_{2},0} - x)}$$ The total pressure at equilibrium is the sum of the partial pressures of both species: $$P = P_{\mathrm{I}_{2}} + P_{\mathrm{I}} = (P_{\mathrm{I}_{2},0} - x) + 2x$$ We also know that \(P = 1.4P_{\mathrm{I}_{2},0}\). Substituting this into the previous equation, we get: $$1.4P_{\mathrm{I}_{2},0} = P_{\mathrm{I}_{2},0} - x + 2x$$

Solve for x

Rearrange the equation from Step 2 to solve for x: $$x = 0.4P_{\mathrm{I}_{2},0}$$ Now that we have determined the value of x, we can use it to find the equilibrium partial pressures of the species.

Determine the equilibrium partial pressures

Using the value of x, we can find the partial pressures at equilibrium: $$P_{\mathrm{I}_{2}} = P_{\mathrm{I}_{2},0} - x = P_{\mathrm{I}_{2},0} - 0.4P_{\mathrm{I}_{2},0} = 0.6P_{\mathrm{I}_{2},0}$$ $$P_{\mathrm{I}} = 2x = 2(0.4P_{\mathrm{I}_{2},0}) = 0.8P_{\mathrm{I}_{2},0}$$

Calculate the value of K

Now we can substitute the equilibrium partial pressures into the expression for K from Step 1: $$K = \frac{(0.8P_{\mathrm{I}_{2},0})^2}{(0.6P_{\mathrm{I}_{2},0})}$$ $$K = \frac{(0.64P_{\mathrm{I}_{2},0})^2}{0.6P_{\mathrm{I}_{2},0}}$$ $$K = \frac{0.64^2}{0.6} = \frac{0.4096}{0.6} \approx 0.683$$ So the equilibrium constant K for this reaction at the given temperature is approximately 0.683.

Key concepts

Chemical Equilibrium

In the fascinating world of chemistry, chemical equilibrium plays a pivotal role. It’s the state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. In simpler words, it’s like a tug-of-war where both teams are equally strong, resulting in no side gaining ground.

When a reaction reaches this point, the concentrations of reactants and products remain constant over time, but don't be fooled – they are not necessarily equal to each other. The reaction is still happening in both directions, but there's no overall change in the amounts of substances. This concept is essential in understanding how reactions proceed and how to manipulate conditions to favor the production of desired compounds.

Partial Pressures

Diving into the realm of gas phase reactions, we encounter the concept of partial pressures. This is basically how much pressure each gas in a mixture would exert if it were alone in the container. Think of it like a group project; each member contributes a certain amount to the final work, that's their 'partial' input.

For gases that follow the ideal gas law, which most do at high temperatures and low pressures, partial pressures become an invaluable tool for calculating reactions involving gases, like in our equilibrium example. By understanding the individual pressures, we can predict how gases will behave and interact in a mixture – a fundamental aspect in fields ranging from chemistry to environmental science.

Equilibrium Constant Expression

Within the framework of chemical equilibrium, the equilibrium constant expression is our mathematical compass. It’s a formula that gives us the equilibrium constant, denoted as K, which quantifies the ratio of concentrations of products to reactants, each raised to the power of their coefficient in the balanced equation.

In the case of gases, we use partial pressures instead of concentrations. Our expression becomes an elegant dance of these pressures. By calculating K, researchers and students can determine how far a reaction will go toward products or reactants under certain conditions, which is crucial for predicting the outcome of chemical processes and designing industrial reactions efficiently.

Gas Phase Reactions

Gas phase reactions are the spectacular showstoppers where reactants in their gaseous forms dance together to form products. These reactions differ from their siblings in liquid or solid phases because the reacting molecules are well separated and move independently, often resulting in different reaction rates and mechanisms.

When studying gas phase reactions, we rely on the ideal gas law and the concept of partial pressures. They help us understand how temperature and pressure affect reaction rates and the position of equilibrium. For example, in our equilibrium exercise, an increased total pressure affects the extent to which iodine dissociates into iodine atoms, illustrating the delicate balance maintained in these reactions.