Consider the following reaction equilibrium
`N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g)`
Initially, 1 mole of `N_(2)` and 3 moles of `H_(2)` are taken in a 2 flask. At equilibrium state if, the number of moles of `N_(2)` is 0.6, what is the total number of moles of all gases present in the flask ?

Options

A. 0.8
B. 1.6
C. 3.2
D. 6.4

Answer

To solve the problem, we will analyze the equilibrium reaction step by step:

**Given Reaction:**
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

**Initial Conditions:**
- Moles of \( N_2 \) = 1 mole
- Moles of \( H_2 \) = 3 moles
- Moles of \( NH_3 \) = 0 moles

**Equilibrium Condition:**
At equilibrium, the moles of \( N_2 \) = 0.6 moles.

**Step 1: Determine the change in moles of \( N_2 \)**
- Initial moles of \( N_2 \) = 1 mole
- Moles of \( N_2 \) at equilibrium = 0.6 moles
- Change in moles of \( N_2 \) = Initial moles - Equilibrium moles
\[ \Delta N_2 = 1 - 0.6 = 0.4 \text{ moles} \]

**Step 2: Determine the change in moles of \( H_2 \)**
From the balanced equation, 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \). Therefore, if 0.4 moles of \( N_2 \) have reacted, the moles of \( H_2 \) that reacted can be calculated as follows:
\[ \Delta H_2 = 3 \times \Delta N_2 = 3 \times 0.4 = 1.2 \text{ moles} \]

**Step 3: Calculate the remaining moles of \( H_2 \)**
- Initial moles of \( H_2 \) = 3 moles
- Moles of \( H_2 \) at equilibrium = Initial moles - Reacted moles
\[ H_2 \text{ at equilibrium} = 3 - 1.2 = 1.8 \text{ moles} \]

**Step 4: Determine the moles of \( NH_3 \) produced**
From the balanced equation, 1 mole of \( N_2 \) produces 2 moles of \( NH_3 \). Therefore, if 0.4 moles of \( N_2 \) have reacted, the moles of \( NH_3 \) produced can be calculated as follows:
\[ \Delta NH_3 = 2 \times \Delta N_2 = 2 \times 0.4 = 0.8 \text{ moles} \]

**Step 5: Calculate the total moles at equilibrium**
Now, we can find the total number of moles of all gases present at equilibrium:
- Moles of \( N_2 \) = 0.6 moles
- Moles of \( H_2 \) = 1.8 moles
- Moles of \( NH_3 \) = 0.8 moles

Total moles at equilibrium:
\[ \text{Total moles} = N_2 + H_2 + NH_3 \]
\[ \text{Total moles} = 0.6 + 1.8 + 0.8 = 3.2 \text{ moles} \]

**Final Answer:**
The total number of moles of all gases present in the flask at equilibrium is **3.2 moles**.

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