The decomposition of `NH_3` on platinum surface is a zero order reaction. What would be the rate of production of `N_2 and H_2` if `k=2.5xx10^(-4) "mol L"^(-1) s^(-1)` ?

Explanation

To solve the problem of the decomposition of \( NH_3 \) on a platinum surface as a zero-order reaction, we can follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition of ammonia (\( NH_3 \)) can be represented as:
\[ 2 NH_3 \rightarrow N_2 + 3 H_2 \]
This indicates that 2 moles of ammonia decompose to produce 1 mole of nitrogen and 3 moles of hydrogen.
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