According to the principle of LeChateller, a higher pressure applied
to the reversible reaction `N_2 + 3H_2 undersetlarr to 2NH_3` would be
expected to result in

Options

A. shifting the equilibrium to the right
B. shifting the equilibrium to the left
C. no change in the equilibrium
D. increased percentages of `NH_3` and `H_2`

Explanation

To solve the problem using Le Chatelier's principle, we can follow these steps:

### Step 1: Identify the Reaction
The given reversible reaction is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

### Step 2: Count the Moles of Gases
- On the left side (reactants), we have:
- 1 mole of \( N_2 \)
- 3 moles of \( H_2 \)

Total moles on the left = \( 1 + 3 = 4 \) moles

- On the right side (products), we have:
- 2 moles of \( NH_3 \)

Total moles on the right = \( 2 \) moles

### Step 3: Apply Le Chatelier's Principle
Le Chatelier's principle states that if an external change is applied to a system at equilibrium, the system will adjust itself to counteract that change and restore a new equilibrium.

In this case, the external change is an increase in pressure.

### Step 4: Determine the Effect of Increased Pressure
When pressure is increased, the equilibrium will shift towards the side with fewer moles of gas.

- Left side has 4 moles (reactants)
- Right side has 2 moles (products)

Since the right side has fewer moles of gas, the equilibrium will shift to the right.

### Step 5: Conclusion
Thus, according to Le Chatelier's principle, applying a higher pressure to the reaction:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
will result in the formation of more \( NH_3 \) as the equilibrium shifts to the right.

### Final Answer
The reaction will shift to the right, favoring the production of \( NH_3 \).

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