Question

sucrose decomposes in acid solution into glucose and fructose according to first order rate law t1/2 = 3hrs.what fraction remains after 9 hrs​

Answer 1

Explanation:

For a first order reaction,

k = 2.303/t Log [R]º / [R]

It is given that, t1/2 = 3.00 hours

Therefore, k = 0.693 / t1/2

= 0.693 / 3 h-1

= 0.231 h - 1

Then, 0.231 h - 1 = 2.303 / 8h Log [R]º / [R]