Decompositon of `NH_(3)` on gold surface follows zero order kinetics. If rate constant is `5xx10^(-4)Ms^(-1)`, rate of formation of `N_(2)` will be

Options

A. `10^(-3)M-s^(-1)`
B. `2.5xx10^(-4)M-s^(-1)`
C. `5xx10^(-4)M-s^(-1)`
D. Zero

Answer

To solve the problem regarding the decomposition of ammonia (NH₃) on a gold surface following zero-order kinetics, we will follow these steps:

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

### Step 2: Write the Rate Law
For a zero-order reaction, the rate of the reaction is independent of the concentration of the reactants. The rate law can be expressed as:
\[ \text{Rate} = k \]
where \( k \) is the rate constant.

### Step 3: Given Data
From the problem, we know:
- The rate constant \( k = 5 \times 10^{-4} \, \text{M/s} \)

### Step 4: Relate the Rate of Formation of Products to the Rate of Reaction
From the stoichiometry of the reaction:
- For every 2 moles of NH₃ decomposed, 1 mole of N₂ is formed.

Thus, the rate of formation of N₂ can be expressed as:
\[ \text{Rate of formation of } \text{N}_2 = \frac{1}{2} \times \text{Rate of disappearance of } \text{NH}_3 \]

### Step 5: Calculate the Rate of Formation of N₂
Since the rate of the reaction is equal to the rate constant for a zero-order reaction:
\[ \text{Rate} = k = 5 \times 10^{-4} \, \text{M/s} \]

Now, substituting this into the equation for the formation of N₂:
\[ \text{Rate of formation of } \text{N}_2 = \frac{1}{2} \times 5 \times 10^{-4} \, \text{M/s} \]

Calculating this gives:
\[ \text{Rate of formation of } \text{N}_2 = \frac{5 \times 10^{-4}}{2} = 2.5 \times 10^{-4} \, \text{M/s} \]

### Final Answer
The rate of formation of \( \text{N}_2 \) is:
\[ \text{Rate of formation of } \text{N}_2 = 2.5 \times 10^{-4} \, \text{M/s} \]

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