Notifications
Clear all
0
30/01/2022 5:00 pm
Topic starter
Consider a certain reaction A → Products with k = 2.0 x 10-2 s-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L-1.
1 Answer
0
30/01/2022 5:15 pm
Given,
k = 2.0 × 10-2 s-1
T = 100 s
[A]o = 1.0 mol-1
Since the unit of k is s-1, the given reaction is a first order reaction.
Therefore, k = \(\frac{2.303}{t}\)log \(\frac{[A]_0}{[A]}\)
⇒ 2.0 x 10-2 s-1 = \(\frac{2.303}{100s}\)log\(\frac{1.0}{[A]}\)
⇒ 2.0 x 10-2 s-1 = \(\frac{2.303}{100s}\)log (-log[A])
⇒ -log[A] = \(\frac{2.0 \times 10^{-2} \times 100}{2.303}\)
⇒ [A] = antilog \(\Big(-\frac{2.0 \times 10^{-2} \times 100}{2.303}\Big)\)
= 0.135 mol L-1 (approximately)
Hence, the remaining concentration of A is 0.135 mol L-1.
This post was modified 3 years ago by admin