Chapter 7: Problem 53
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
Step by step solution
Write the Equilibrium Expression
Determine Initial Concentrations
Write Equilibrium Concentrations in terms of x
Substitute into the Equilibrium Expression
Solve the Equation for x
Calculate the Mole Fraction of R
Identify the Correct Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mole Fraction
To calculate the mole fraction of a component like \([]R[]\), we use the formula:
- \( \text{Mole fraction of } R = \frac{[R]}{[R] + [P] + [Q]} \)
Equilibrium Expression
For the given reaction, the expression is set up based on the equilibrium constant \(K_{eq}\), expressed as:
- \[ K_{eq} = \frac{[R]^2}{[P]^{81}[Q]} \]
Initial Concentrations
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: \([P] = 1 - 81x\),
- \([Q] = 1 - x\),
- \([R] = 2x\)
Reaction Shift
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:
- \(1 - 81x \approx 1\)