Question

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with ${t_{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
 {\vphantom {1 2}}\right.}
\!\lower0.7ex\hbox{$2$}}}} = 3.00$ hours. What fraction of samples of sucrose remains after $8$ hours?

Answer 1

Hint:Use the relation between half-life and rate constant of the reaction. From the given value of half-life find the value of rate constant. After this, use the formula of rate constant for a first order reaction and calculate the amount of sample that is unreacted after $8$ hours.

Complete step-by-step solution:
Given that the reaction is of first order and the half-life is given as ${t_{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
 {\vphantom {1 2}}\right.}
\!\lower0.7ex\hbox{$2$}}}} = 3.00$ hours.
For a first order reaction, we can write the relation between half-life and rate constant as $k = \dfrac{{0.693}}{{{t_{{1 \mathord{\left/
 {\vphantom {1 2}} \right.
 } 2}}}}}$
Now, we substitute the given value of half-life. $k = \dfrac{{0.693}}{{3hr}} = 0.231h{r^{ - 1}}$
Now, let ${C_0}$ be the initial concentration of sucrose and ${C_t}$ be the concentration of sucrose after time t.
The formula for rate constant is $k = \dfrac{{2.303}}{t}\log \dfrac{{{C_0}}}{{{C_t}}}$
We can now substitute the values of time and rate constant to calculate the ratio between ${C_0}$ and ${C_t}$
$0.231 = \dfrac{{2.303}}{8}\log \dfrac{{{C_0}}}{{{C_t}}}$
$\Rightarrow \log \dfrac{{{C_o}}}{{{C_t}}} = \dfrac{{0.231 \times 8}}{{2.303}}$
By solving this, we get $\log \dfrac{{{C_0}}}{{{C_t}}} = 0.8024$
$\dfrac{{{C_0}}}{{{C_t}}} = anti\log (0.8024) = 6.3445$
Now, to calculate the fraction we take the reciprocal of the acquired value which is
$\dfrac{{{C_t}}}{{{C_0}}} = \dfrac{1}{{6.3445}} = 0.15761 \simeq 0.158$

Therefore, the fraction of sucrose that remained unreacted after $8$ hours is $0.158$.

Note: In order to calculate the amount of unreacted sucrose we have to first find the rate constant for a first order reaction. After calculating the rate constant, we use the formula for rate constant to derive a relation between initial concentration of sucrose and the concentration left after $8$ hours. This relation gave us the fraction of sucrose that was left.