The decomposition of `NH_(3)` on platinum surface is zero order reaction the rate of production of `H_(2)` is `(k=2.5xx10^(-4) M s^(-1))`

Options

A. `2.5xx10^(-4) M s^(-1)`
B. `1.25 xx 10^(-4) M s^(-1)`
C. `3.75 xx10^(-4) M s^(-1)`
D. `5.0 xx10^(-4) M s^(-1)`

Explanation

To solve the problem regarding the decomposition of ammonia (NH₃) on a platinum surface, we need to follow these steps:

### Step 1: Understand the Reaction
The decomposition of ammonia can be represented by the following balanced chemical equation:
\[ 2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2 \]

### Step 2: Identify the Order of Reaction
The problem states that the decomposition of NH₃ is a zero-order reaction. In a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of the reactants.

### Step 3: Write the Rate Law
For a zero-order reaction, the rate law can be expressed as:
\[ R = k \]
Where:
- \( R \) is the rate of the reaction,
- \( k \) is the rate constant.

Given that \( k = 2.5 \times 10^{-4} \, \text{M s}^{-1} \), we can state:
\[ R = 2.5 \times 10^{-4} \, \text{M s}^{-1} \]

### Step 4: Relate the Rate of Production of H₂ to the Rate of Reaction
From the balanced equation, we see that 3 moles of H₂ are produced for every 2 moles of NH₃ decomposed. Thus, the rate of formation of H₂ can be related to the rate of the reaction:
\[ R_{\text{H}_2} = \frac{3}{2} R \]

### Step 5: Calculate the Rate of Production of H₂
Substituting the value of \( R \) into the equation gives:
\[ R_{\text{H}_2} = \frac{3}{2} (2.5 \times 10^{-4}) \]
\[ R_{\text{H}_2} = \frac{3 \times 2.5}{2} \times 10^{-4} \]
\[ R_{\text{H}_2} = \frac{7.5}{2} \times 10^{-4} \]
\[ R_{\text{H}_2} = 3.75 \times 10^{-4} \, \text{M s}^{-1} \]

### Final Answer
The rate of production of H₂ is:
\[ R_{\text{H}_2} = 3.75 \times 10^{-4} \, \text{M s}^{-1} \]

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