For a reversible gaseous reaction `N_(2)+3H_(2)hArr2NH_(3)` at equilibrium , if some moles of `H_(2)` are replaced by same number of moles of `T_(2)` (T is tritium , isotope of H and assume isotopes do not have different chemical properties ) without affecting other parameters , then:

Options

A. the sample of ammonia obtained after something will be radioactive .
B. moles of `N_(2)` after the change will be different as compared to moles of `N)(2)` present before the change
C. the volue of `K_(p) "or" K_(c)` will change
D. the average molecular mass of new equilibrium will be same as that of old equilibrium

Answer

To solve the problem, we need to analyze the effects of replacing some moles of \( H_2 \) with the same number of moles of tritium (\( T_2 \)) in the equilibrium reaction:

\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction involves nitrogen gas (\( N_2 \)) reacting with hydrogen gas (\( H_2 \)) to form ammonia (\( NH_3 \)). The stoichiometry shows that 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \) to produce 2 moles of \( NH_3 \).

2. **Replacement of Hydrogen with Tritium**:
When some moles of \( H_2 \) are replaced by tritium (\( T_2 \)), we assume that tritium behaves chemically the same as hydrogen. Therefore, if we replace \( x \) moles of \( H_2 \) with \( x \) moles of \( T_2 \), the reaction can still proceed as follows:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
can also produce:
\[ N_2 + 3T_2 \rightleftharpoons 2NH_3 \]
and mixed products like \( NH_2T \) or \( NHT_2 \).

3. **Formation of Radioactive Products**:
Since tritium is radioactive, any ammonia produced that contains tritium will also be radioactive. Therefore, if we have replaced some hydrogen with tritium, the resulting ammonia will contain tritium isotopes, making it radioactive.

4. **Effect on Equilibrium Concentrations**:
The moles of \( N_2 \) will remain unchanged because the replacement of \( H_2 \) with \( T_2 \) does not affect the nitrogen. The equilibrium constant \( K_p \) or \( K_c \) will also remain unchanged because the chemical properties of the isotopes are the same, and the reaction conditions are not altered.

5. **Average Molecular Mass**:
The average molecular mass of the products will change because tritium has a different mass than hydrogen. Therefore, the average molecular mass of the new equilibrium will not be the same as that of the old equilibrium.

### Conclusion:
From the analysis, we can conclude that:

- The sample of ammonia obtained after some time will be radioactive due to the presence of tritium.
- The moles of \( N_2 \) remain unchanged.
- The equilibrium constant \( K_p \) or \( K_c \) remains unchanged.
- The average molecular mass of the new equilibrium will be different from that of the old equilibrium.

Thus, the correct answer is that the sample of ammonia obtained after some time will be radioactive.