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# Solution - 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? - Integrated Rate Equations - Half-life of a Reaction

ConceptIntegrated Rate Equations Half-life of a Reaction

#### 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?

#### 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

2014-2015 (March)
Question 3.3 | 3 marks

#### Video TutorialsVIEW ALL [1]

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 (Computer Science), HSC Science (Electronics), HSC Science (General)
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