#### Question

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with t_{1/2}= 3 hours. What fraction of the sample of sucrose remains after 8 hours?

clickto share

#### Solution

`t_(1/2)=3 Hours`

Now we know that,

`k=0.693/t_(1/2)=0.693/3=0.231 h^(-1)`

Put above value in the formula of first order reaction,

`k=2.303/t log([R]_0/([R]))=2.303/8log([R]_0/([R]))`

So,

`log([R]_0/([R]))=(0.231xx8)/2.303=0.8024`

Taking antilog on both sides,

`[R]_0/([R])=6.3445`

`[R]/[R]_0=0.158`

Fraction of the sample of sucrose remaining after 8 hours = 0.158

Is there an error in this question or solution?

#### APPEARS IN

Solution for question: Sucrose Decomposes in Acid Solution into Glucose and Fructose According to the First Order Rate Law with T1/2= 3 Hours. What Fraction of the Sample of Sucrose Remains After 8 Hours? concept: Integrated Rate Equations - Half-life of a Reaction. For the courses HSC Science (General) , HSC Science (Electronics), HSC Science (Computer Science)