Sucrose hydrolyses in acidic medium to form glucose and fructose and follows first-order kinetics.

If the half-life of sucrose is 3 hours, find the percentage of sucrose left after 6 hours.

For a first-order reaction:

t = n × t_1/2

Given:

• Half-life = 3 hours
• Total time = 6 hours

Number of half-lives:

n = 6/3 = 2

After one half-life:

Remaining amount = 1/2

After two half-lives:

Remaining amount = (1/2)²

= 1/4

Percentage remaining:

= 25%

Thus, after 6 hours only one-fourth of the original sucrose remains unreacted.

Final Answer

25%