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?
(\(k = 0.5 × 10^{-4}\text{ mol L}^{-1}s^{-1}\))
Answer & explanation
Correct answer: option 1
The correct answer is option 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,
\(t_{1/2} = \frac{[R]_0}{2K}\)
\(⇒ t_{1/2} = \frac{1 × 10^{-4}}{2 × 0.5 × 10^{-4}}\)
\(⇒ t_{1/2} = \frac{1 × 10^{-4}}{1 × 10^{-4}}\)
\(⇒ t_{1/2} = 1s\)