Question

One mole of a compound AB reacts with one mole of compound CD according to the equation: \(\mathrm{AB}+\mathrm{CD} \rightarrow \mathrm{AD}+\mathrm{CB}\) When equilibrium had been established it was found that \(3 / 4\) mole each of reactant \(\mathrm{AB}\) and \(\mathrm{CD}\) has been converted to AD and CB. There is no change in volume. The equilibrium constant of the reaction is? (a) 9 (b) \(\frac{1}{9}\) (c) \(9 / 16\) (d) \(16 / 9\)

Short answer

The equilibrium constant is 9.

Step-by-step solution

Understanding Initial Conditions

Initially, you have 1 mole each of AB and CD. Therefore, the initial concentration of AB and CD is each 1 mole per liter, assuming 1-liter volume.

Calculating Changes at Equilibrium

At equilibrium, it is given that 3/4 mole each of AB and CD has reacted. Thus, the concentration of AB and CD each decreases by 3/4 mole, leaving 1 - 3/4 = 1/4 mole of each at equilibrium.

Concentration of Products at Equilibrium

Since 3/4 mole of each reactant reacts, it forms 3/4 mole of each product, AD and CB. Therefore, the equilibrium concentration of AD and CB is 3/4 mole per liter.

Writing the Expression for Equilibrium Constant

The equilibrium constant expression for the reaction \( \mathrm{AB} + \mathrm{CD} \rightarrow \mathrm{AD} + \mathrm{CB} \) is given by: \[ K_c = \frac{[\mathrm{AD}][\mathrm{CB}]}{[\mathrm{AB}][\mathrm{CD}]} \] Substitute the concentrations at equilibrium into this expression.

Substituting Equilibrium Concentrations

Plug the known concentrations into the equilibrium expression: \[ K_c = \frac{(3/4)(3/4)}{(1/4)(1/4)} \] Simplify and calculate.

Calculating the Equilibrium Constant

Calculate the values: \[ K_c = \frac{9/16}{1/16} = 9 \]. Therefore, the equilibrium constant \( K_c \) is 9.

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs in a reversible chemical reaction when the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant yet are not necessarily equal. The reaction does not stop; instead, it proceeds indefinitely, where the forward and backward processes continue at the same rate. This creates a balance allowing specific proportions of substances, characteristic to the equilibrium state, to be maintained.

In the exercise, once the chemical system involving the reaction between AB and CD reaches equilibrium, the conversion of reactants to products proceeds at the same speed as their reconversion into reactants. Understanding equilibrium is vital for controlling reaction conditions and predicting the concentration of substances needed to achieve desired results.

Mole Calculations

Mole calculations are fundamental in chemistry, providing a method to express the amount of a substance. A mole represents Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles of the substance. In this context, moles allow chemists to translate between the microscopic scale of atoms and molecules to the macroscopic amounts they measure in the lab.

• The initial problem provides 1 mole of AB and 1 mole of CD, which simplifies calculations, allowing a straightforward determination of changes in quantities.
• As 3/4 mole of AB and CD are transformed, these calculations involve subtracting and adding mole quantities to determine both reactant and product concentrations at equilibrium.

Always remember: maintaining proper mole calculations ensures accuracy when predicting and measuring chemical reactions.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. Stoichiometry uses balanced chemical equations to ensure the correct proportions are maintained. This relationship guides how much of each reactant is required, and how much product will form.

In our provided reaction, \( \mathrm{AB} + \mathrm{CD} \rightarrow \mathrm{AD} + \mathrm{CB} \), stoichiometry is essential for understanding how 1 mole of AB reacts with 1 mole of CD to form 1 mole each of AD and CB. This relationship is shown by the balanced equation, which helps predict the amounts of product formed from given amounts of reactants.

• The calculated equilibrium concentrations stem directly from these stoichiometric relationships.
• By comprehending these connections, you can successfully navigate mole changes in both laboratory settings and theoretical exercises.

Equilibrium Concentrations

At equilibrium, the concentrations of reactants and products have settled into a steady state. These concentrations depend on reactions' specifics and initial conditions. Knowing the equilibrium constant, \( K_c \), and substituting equilibrium concentrations helps predict the direction and extent of reactions.

For AB and CD, the equilibrium concentrations were determined by noting that 3/4 of each had reacted, leaving 1/4 mole of reactants, while forming 3/4 mole of products, AD and CB.

The equilibrium constant expression, \( K_c = \frac{[\mathrm{AD}][\mathrm{CB}]}{[\mathrm{AB}][\mathrm{CD}]} \), involves these concentrations. You substitute and calculate values to find \( K_c \), simplifying to give a numeric value, in this case 9.

• Understanding equilibrium concentrations helps anticipate how changes can shift the equilibrium position or impact \( K_c \).
• This knowledge aids in chemical manufacturing, environmental science, and various strategic applications utilizing chemical processes.