Read the following paragraph and answer the questions given below,
for a general reaction, `aA+bB hArr cC=dD`, equilibrium constant `K_c` is given by `K_c=([C]^c[D]^d)/([A]^a[B]^b)` However, when all reactants and products are gases the equilibrium constant is generally expressed in terms of partial pressure.
`K_p=(P_C^cxx P _D^d)/(P_A^axxP_B^b)`
If in reaction `N_2O_4 hArr 2NO_2 , alpha` is the part of `N_2O_4` which dissociates then the number of moleculars at equilibrium to the initial molecules will be

Options

A. 1
B. `1-alpha`
C. `2alpha`
D. `1+alpha`

Answer

To solve the problem, we need to analyze the dissociation of the given reaction and determine the ratio of the number of molecules at equilibrium to the initial number of molecules.

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction given is:
\[ N_2O_4 \rightleftharpoons 2NO_2 \]

2. **Define Initial Conditions**:
Let's assume we start with 1 mole of \( N_2O_4 \) at the beginning (initially). Therefore, the initial number of molecules is:
\[ \text{Initial molecules} = 1 \]

3. **Define the Dissociation**:
Let \( \alpha \) be the fraction of \( N_2O_4 \) that dissociates. Therefore, the amount of \( N_2O_4 \) that dissociates is:
\[ \text{Dissociated} = \alpha \]
Hence, the amount of \( N_2O_4 \) remaining at equilibrium will be:
\[ \text{Remaining } N_2O_4 = 1 - \alpha \]

4. **Calculate the Amount of \( NO_2 \) Produced**:
According to the stoichiometry of the reaction, for every mole of \( N_2O_4 \) that dissociates, 2 moles of \( NO_2 \) are produced. Therefore, the amount of \( NO_2 \) produced at equilibrium is:
\[ \text{Produced } NO_2 = 2\alpha \]

5. **Total Number of Molecules at Equilibrium**:
The total number of molecules at equilibrium will be the sum of the remaining \( N_2O_4 \) and the produced \( NO_2 \):
\[ \text{Total molecules at equilibrium} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

6. **Calculate the Ratio**:
Now, we need to find the ratio of the number of molecules at equilibrium to the initial number of molecules:
\[ \text{Ratio} = \frac{\text{Total molecules at equilibrium}}{\text{Initial molecules}} = \frac{1 + \alpha}{1} \]
Therefore, the ratio is:
\[ \text{Ratio} = 1 + \alpha \]

### Final Answer:
The number of molecules at equilibrium to the initial molecules is \( 1 + \alpha \).