The decomposition of ammonia on platinum surface is a zero order reaction. How much time it will take for \(1 × 10^{-4} \text{mol L}^{-1}\) of ammonia to reduce into half of its concentration?

(\(k = 0.5 × 10^{-4}\text{ mol L}^{-1}s^{-1}\))

1 s

Given,

\(k = 0.5 × 10^{-4}\text{ mol L}^{-1}s^{-1}\)

Initial concentration, \([R]_0 = 1 × 10^{-4}\text{ mol L}^{-1}\)

Final concentration, \([R]_t = \frac{[R]_0}{2} = \frac{1 × 10^{-4}\text{ mol L}^{-1}}{2}\)

We know, that for a zero-order reaction,

\(t_{1/2} = \frac{[R]_0}{2K}\)

\(⇒ t_{1/2} = \frac{1 × 10^{-4}}{2 × 0.5 × 10^{-4}}\)

\(⇒ t_{1/2} = \frac{1 × 10^{-4}}{1 × 10^{-4}}\)

\(⇒ t_{1/2} = 1s\)