The decomposition of `NH_(3)` on Pt surface is a zero order reaction. If the value of rate constant is `2 xx 10^(-4)" mole liter"^(-1)" sec"^(-1)` . The rate of appearance of `N_(2)` and `H_(2)` are respectively
`{:(" N"_(2)," H"_(2)),("(1) "1xx10^(-4)" mol l"^(-1)"sec"^(-1),3xx10^(-4)" mol"^(-1)"sec"^(-1)),("(2) "3xx10^(-4)" mol l"^(-1)"sec"^(-1),1xx10^(-4)" mol"^(-1)"sec"^(-1)),("(3) "2xx10^(-4)" mol l"^(-1)"sec"^(-1),6xx10^(-4)" mol"^(-1)"sec"^(-1)),("(4) "3xx10^(-4)" mol l"^(-1)"sec"^(-1),3xx10^(-4)" mol"^(-1)"sec"^(-1)):}`

Answer

To solve the problem, we need to analyze the decomposition of ammonia (NH₃) on a platinum surface, which is given as a zero-order reaction. The reaction can be represented as follows:

\[ 2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2 \]

### Step 1: Identify the Rate of Reaction
For a zero-order reaction, the rate of reaction is constant and is equal to the rate constant \( k \). Given that \( k = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \), we can express the rate of appearance of products in terms of \( k \).

### Step 2: Relate the Rates of Appearance of Products
From the balanced equation, we can relate the rates of appearance of nitrogen (\( \text{N}_2 \)) and hydrogen (\( \text{H}_2 \)):

\[ -\frac{1}{2} \frac{d[\text{NH}_3]}{dt} = \frac{d[\text{N}_2]}{dt} = \frac{1}{3} \frac{d[\text{H}_2]}{dt} \]

### Step 3: Express the Rate of Formation of \( \text{N}_2 \)
From the equation, we can express the rate of formation of \( \text{N}_2 \):

\[ \frac{d[\text{N}_2]}{dt} = k = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 4: Express the Rate of Formation of \( \text{H}_2 \)
Using the stoichiometric relationship, we can find the rate of formation of \( \text{H}_2 \):

\[ \frac{d[\text{H}_2]}{dt} = 3 \times \frac{d[\text{N}_2]}{dt} = 3 \times (2 \times 10^{-4}) = 6 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 5: Compile the Results
Now we have the rates of appearance for both products:
- Rate of appearance of \( \text{N}_2 \): \( 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)
- Rate of appearance of \( \text{H}_2 \): \( 6 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)

### Conclusion
The correct option that matches these rates is:

**Option (3): \( (2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1}, 6 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1}) \)**

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