#### 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?

#### 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

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#### Related Questions VIEW ALL [4]

The experimental data for decomposition of N_{2}O_{5}

[`2N_2O_5 -> 4NO_2 + O_2`] in gas phase at 318K are given below:

t(s |
0 | 400 | 800 | 1200 | 1600 | 2000 | 2400 | 2800 | 3200 |

`10^2xx[N_2O_5]mol L^(-1)` | 1.63 | 1.36 | 1.14 | 0.93 | 0.78 | 0.64 | 0.53 | 0.43 | 0.35 |

**(i) **Plot [N_{2}O_{5}] against *t*.

**(ii) **Find the half-life period for the reaction.

**(iii)** Draw a graph between log [N_{2}O_{5}] and *t.*

**(iv) **What is the rate law?

**(v) **Calculate the rate constant.

**(vi) **Calculate the half-life period from *k *and compare it with (ii).