Question

If in the reaction \(N_2O_4 \rightleftharpoons 2NO_2\), \( \alpha \) is the degree of dissociation of \(N_2O_4\), then total number of moles at equilibrium is:

• A. \(1 - \alpha\)
• B. \(1 + \alpha\)
• C. \(1 + 2\alpha\)
• D. \(1 + \frac{\alpha}{2}\)

Hint:

Total moles = initial + change due to dissociation.

Answer 1

The Correct Option is B

To solve the question of determining the total number of moles at equilibrium in the reaction \(N_2O_4 \rightleftharpoons 2NO_2\) given the degree of dissociation \(\alpha\), we can follow these steps:

1. Identify the initial number of moles of \(N_2O_4\). Assume the initial moles of \(N_2O_4\) is 1 mole at the start of the reaction.
2. Understand the concept of degree of dissociation. \(\alpha\) represents the fraction of the initial amount of a substance that has dissociated. For \(N_2O_4\), \(\alpha\) is the degree of dissociation, meaning that a fraction \(\alpha\) of the initial 1 mole of \(N_2O_4\) dissociates into \(NO_2\).
3. Determine the change in moles during the reaction. For every mole of \(N_2O_4\) that dissociates, 2 moles of \(NO_2\) are produced. Therefore, if \(\alpha\) moles of \(N_2O_4\) dissociate, then \(2\alpha\) moles of \(NO_2\) are formed.
4. Calculate the moles of substances at equilibrium:
   • Moles of \(N_2O_4\) remaining at equilibrium = \(1 - \alpha\)
   • Moles of \(NO_2\) formed at equilibrium = \(2\alpha\)
5. The total number of moles at equilibrium is the sum of the moles of all substances present:
   • Total moles at equilibrium = \((1 - \alpha) + 2\alpha = 1 + \alpha\)

Therefore, the correct option that represents the total number of moles at equilibrium is \(1 + \alpha\), which is Option: \(1 + \alpha\).