Question

3\. If \(K_{e q}\) for the reaction is \(81 \mathrm{P}+\mathrm{Q} \rightleftharpoons 2 \mathrm{R}\) If we start with 1 mole each of \(\mathrm{P}\) and \(\mathrm{Q} .\) What is the mole fraction of \(\mathrm{R}\) at equilibrium: (a) \(\frac{1}{9}\) (b) \(\frac{11}{9}\) (c) \(\frac{4}{9}\) (d) \(\frac{9}{11}\)

Short answer

(d) \(\frac{9}{11}\)

Step-by-step solution

Write the Equilibrium Expression

For the reaction \(81 \mathrm{P} + \mathrm{Q} \rightleftharpoons 2 \mathrm{R}\), the equilibrium constant \(K_{eq}\) is given by: \[ K_{eq} = \frac{[R]^2}{[P]^{81}[Q]} \]Given that \(K_{eq} = 81\).

Determine Initial Concentrations

Initially, \([P] = 1\) mole and \([Q] = 1\) mole. Let the change at equilibrium be \(x\) such that \([R] = 2x\) at equilibrium since 2 moles of \(R\) form for 1 mole each of \(P\) and \(Q\) reacting.

Write Equilibrium Concentrations in terms of x

At equilibrium, \([P] = 1 - 81x\), \([Q] = 1 - x\),\([R] = 2x\).These represent the concentrations of each species in terms of \(x\).

Substitute into the Equilibrium Expression

Using the equilibrium expression:\[81 = \frac{(2x)^2}{(1-81x)^{81}(1-x)}\]Solve this equation for \(x\).

Solve the Equation for x

Assuming a small extent of reaction due to large power, approximate:\[1 - 81x \approx 1\]thus,\[81 \approx \frac{4x^2}{1-x}\]By simplifying further, approximate when solving for small \(x\).

Calculate the Mole Fraction of R

Substituting approximate values:Let \(x \approx \frac{1}{11}\) based on small reaction shift due to large \(K_{eq}\). Calculate \([R] = 2x = \frac{2}{11} \), substitute this into \[\text{Mole fraction of } R = \frac{[R]}{[R]+[P]+[Q]} = \frac{\frac{2}{11}}{\frac{2}{11}+(1-\frac{81x}{11})+(1-\frac{x}{11})} \]Simplify for small \(x\), \[\text{Mole fraction of } R \approx \frac{9}{11}\].

Identify the Correct Answer

After calculations, the mole fraction of \(R\) at equilibrium is \(\frac{9}{11}\) which corresponds to option (d).

Key concepts

Mole Fraction

The concept of mole fraction is crucial when dealing with mixtures of substances, as it refers to the ratio of the number of moles of one component to the total number of moles of all components in the mixture. In a chemical reaction setting, it helps us understand how much of each substance is present at equilibrium.
To calculate the mole fraction of a component like \([]R[]\), we use the formula:

• \( \text{Mole fraction of } R = \frac{[R]}{[R] + [P] + [Q]} \)

In this reaction setup, the mole fraction gives insights into the composition of the reaction mixture at equilibrium, allowing us to predict the concentration of products like \([]R[]\) as compared to the reactants \([]P[]\) and \([]Q[]\). Keeping track of mole fractions is essential in chemical processes and analyses as it provides a dimensionless number that simplifies calculations and comparisons.

Equilibrium Expression

The equilibrium expression is a mathematical representation that helps determine the concentrations of reactants and products in a chemical reaction once equilibrium is achieved.
For the given reaction, the expression is set up based on the equilibrium constant \(K_{eq}\), expressed as:

• \[ K_{eq} = \frac{[R]^2}{[P]^{81}[Q]} \]

This expression relates the equilibrium concentrations of products and reactants through the law of mass action. The proclivity of the reaction to produce products over reactants is signified by the size of \(K_{eq}\). A large \(K_{eq}\), like 81 in this case, hints that the reaction strongly favors the formation of products, influencing the reaction equilibrium position significantly.

Initial Concentrations

Understanding initial concentrations is vital as it lays the groundwork for analyzing how a chemical reaction progresses towards equilibrium. Here, we start with \([]P[] = 1\) mole and \([]Q[] = 1\) mole, setting a basis for determining changes as the reaction advances.
The change is often expressed in terms of a variable \(x\), which represents the amount of reactant converted to product. As the reaction progresses, it subtracts from initial concentrations, influencing how we model equilibria:

• At equilibrium: \([P] = 1 - 81x\),
• \([Q] = 1 - x\),
• \([R] = 2x\)

Initial concentrations allow us to assess reaction extent and the position of equilibrium, impacting the concentration of substances like \([]R[]\) at equilibrium.

Reaction Shift

Reaction shift describes how the concentration of reactants and products change as a reaction progresses towards equilibrium. In this scenario, we examine shifts to predict the final concentration of substances involved, using approximations and simplifications based on initial conditions and equilibrium constants.
The large exponent \(81\) in \([P]^{81}\) suggests small shifts due to significant reactant influence. This indicates a minor extent of reaction, allowing assumptions such as:

• \(1 - 81x \approx 1\)

Here, solving \(81 \approx \frac{4x^2}{1-x}\) provides \(x\) values that reveal minor adjustments toward equilibrium. Understanding these shifts is crucial for approximating the mole fraction of products like \([]R[]\), guiding predictions in chemical equilibria.