At high pressure the following reaction is zero order.
`" "2NH_(3)(g) underset("Platinum catalyst")overset(1130 K)to N_(2)(g) + 3H_(2)(g)`
Which of the following options are correct for this reaction ?

Options

A. Rate of reaction = Rate constant
B. Rate of the reaction depends on concentration of ammonia
C. Rate of decomposition of ammonia will remin constant until ammonia disappears completely
D. Further increases in pressure will change the ratio of reaction

Explanation

To analyze the given reaction and determine which options are correct, we will go through the properties of zero-order reactions and apply them to the provided reaction.

### Step-by-Step Solution:

1. **Identify the Reaction Type**:
The reaction given is:
\[ 2NH_3(g) \xrightarrow{\text{Platinum catalyst, 1130 K}} N_2(g) + 3H_2(g) \]
It is stated that at high pressure, this reaction behaves as a zero-order reaction.

2. **Understanding Zero-Order Reactions**:
In a zero-order reaction, the rate of reaction is independent of the concentration of the reactants. The rate can be expressed as:
\[ \text{Rate} = k \]
where \( k \) is the rate constant.

3. **Evaluate the Options**:
- **Option 1**: "Rate of reaction is equal to rate constant."
- Since the reaction is zero-order, the rate of reaction is indeed equal to the rate constant \( k \).
- **Conclusion**: This option is **correct**.

- **Option 2**: "Rate of reaction depends on concentration of ammonia."
- For a zero-order reaction, the rate does not depend on the concentration of the reactants, including ammonia.
- **Conclusion**: This option is **incorrect**.

- **Option 3**: "Rate of decomposition of ammonia will remain constant until ammonia disappears completely."
- In a zero-order reaction, the rate remains constant until the reactant is completely consumed. Thus, the rate of decomposition of ammonia will remain constant.
- **Conclusion**: This option is **correct**.

- **Option 4**: "Further increases in pressure will change the ratio of the reaction."
- According to Le Chatelier's principle, increasing the pressure in a gaseous reaction will shift the equilibrium towards the side with fewer moles of gas. In this case, the reaction has 4 moles of gas on the left (2 NH3) and 4 moles on the right (1 N2 + 3 H2). Thus, increasing pressure will affect the equilibrium position.
- **Conclusion**: This option is **correct**.

4. **Final Summary of Options**:
- **Correct Options**: 1, 3, and 4.
- **Incorrect Option**: 2.