Question

At \(627^{\circ} \mathrm{C}, K=0.76\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ Calculate \(K\) at \(627^{\circ} \mathrm{C}\) for (a) the synthesis of one mole of sulfur trioxide gas. (b) the decomposition of two moles of \(\mathrm{SO}_{3}\).

Short answer

Question: Calculate the equilibrium constants for the following derivative reactions, given the original reaction and K value: Original reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), K = 0.76 Derivative reactions: (a) SO₂(g) + 1/2O₂(g) ⇌ SO₃(g) (b) 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) Answer: (a) For the synthesis of one mole of sulfur trioxide, K = 0.87. (b) For the decomposition of two moles of sulfur trioxide, K = 1.32.

Step-by-step solution

(Step 1: Identify the derivative reactions)

(The derivative reactions are: (a) Synthesis of one mole of sulfur trioxide: $$\mathrm{SO}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)$$ (b) Decomposition of two moles of sulfur trioxide: $$2\mathrm{SO}_{3}(g) \rightleftharpoons 2\mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g)$$)

(Step 2: Manipulate the original reaction to find the derivative reaction)

(Compare the derivative reactions with the original reaction to determine the relationship between them: (a) The synthesis of one mole of sulfur trioxide is half of the original reaction. (b) The decomposition of two moles of sulfur trioxide is the reverse of the original reaction with no change in stoichiometry.)

(Step 3: Apply the rules for manipulating equilibrium constants)

(Using the relationships identified in Step 2, apply the rules for manipulating equilibrium constants: (a) If a reaction is halved, the equilibrium constant should be squared: $$K_{1} =\sqrt{K}$$ (b) If a reaction is reversed, the equilibrium constant should be inverted: $$K_{2} = \frac{1}{K}$$)

(Step 4: Calculate the new equilibrium constants)

(Using the original K value of 0.76, calculate the new equilibrium constants: (a) Synthesis of one mole of sulfur trioxide: $$K_{1} =\sqrt{0.76} = 0.87$$ (b) Decomposition of two moles of sulfur trioxide: $$K_{2} = \frac{1}{0.76} = 1.32$$) The equilibrium constants for the two derivative reactions are: (a) For the synthesis of one mole of sulfur trioxide, K = 0.87. (b) For the decomposition of two moles of sulfur trioxide, K = 1.32.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating concept in chemistry where a reversible reaction reaches a state where the rate of the forward reaction equals the rate of the reverse reaction. This balance means that the concentrations of the reactants and products remain constant over time. It's important to understand that equilibrium does not mean that the reactants and products are in equal concentrations, but rather that their concentrations do not change as the system is at rest.

Several factors can influence chemical equilibrium, such as temperature, pressure, and concentration of reactants or products. When any of these conditions change, the equilibrium can shift according to Le Chatelier's Principle, which states that the system will adjust to counteract the change and re-establish equilibrium.

In the case of our example with sulfur dioxide and sulfur trioxide, at a temperature of 627°C, the given equilibrium constant, K, helps to quantify the position of the equilibrium and predict the reaction's behavior under these conditions.

Reversible Reactions

Reversible reactions are chemical reactions where the reactants form products, which can themselves react to give the original reactants back. This ongoing process happens until equilibrium is achieved, where the rates of the forward and backward reactions become equal.

For instance, consider the reaction between sulfur dioxide and oxygen to form sulfur trioxide:

• Forward reaction: \[ 2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) \]

The equilibrium nature of the reaction is represented by the double-headed arrow. Such reactions are the backbone of many industrial processes where the formation and breakdown of products occur simultaneously, like the Haber process for ammonia production.

Understanding reversible reactions is crucial for manipulating conditions to favor the desired products, thereby optimizing yields in chemical manufacturing. It's fascinating how altering simple variables like temperature or pressure can tip the scales of equilibrium to favor different directions of the reaction.

Equilibrium Calculations

Equilibrium calculations are a vital part of quantifying how far a reaction has proceeded and understanding the extent of product formation at equilibrium. These calculations typically involve the equilibrium constant, K, which is derived from the concentrations of products and reactants at equilibrium.

The reaction provided earlier has an equilibrium constant K = 0.76 at 627°C for the formation of sulfur trioxide from sulfur dioxide and oxygen. When calculating the equilibrium constant for derivative reactions, certain rules apply:

• If the reaction equation is altered, the equilibrium constant undergoes specific transformations.
• If a reaction is halved, as in the case of forming one mole of \( \mathrm{SO}_{3} \), the equilibrium constant is squared. Resulting in \( K_{1} = \sqrt{K} = 0.87 \).
• If the reaction is reversed, as for the decomposition of two moles of \( \mathrm{SO}_{3} \), the equilibrium constant is inverted, yielding \( K_{2} = \frac{1}{K} = 1.32 \).

These calculations show how changes in the stoichiometry or direction of the reaction affect the equilibrium constant, crucial for predicting reaction behavior under different conditions.