chemistry help required please.....

Question

In an enzyme solution ,sucrose undergoes fermentation. if  0.10 M solution of sucrose is
reduced to 0.05 M in 10 hours  and to 0.025  M in 20 hours , the rate constant of the reaction
is?

    

•  2.5  x 10^-5     s^-1
•  1.9  x 10^-5     s^-1
•  1.7 x 10^-5     mol dm^-3  s^-1
•  2.0 x 10^-5     mol dm^-3 s^-1
  
   

   

  although i have tried  to use the formula  rate =  (concentration final - concentration initial )/ time (final ) - time (initial)

  i am not sure whether it is correct or not ,even the answer  is not coming .

Answer 1

The first thing to do is figure out the order of the reaction. Most common chemical reactions are either first or zeroth order reactions. To distinguish between the two:

 

- Zeroth order reactions consume a constant AMOUNT of reactant and make a constant AMOUNT of product per unit time. 

- First order reactions consume a constant FRACTION of reactant and make a constant FRACTION of product per unit time. 

 

Looking at this question, you are given the amount of a reactant (sucrose) at different time points (every 10 hours). You can easily see that the amount drops by half every 10 hours which suggests that the reaction is a first order reaction. 

 

Since the fraction of reactant consumed per unit time is constant in first order reactions, that should suggest that if you plotted reactant concentration over time you should get an exponential decay and that the rate equation should involve an exponential term.

 

In fact the rate equation for first order reactions is: [S] = [S]₀e^-kt

 

where [S] is the reactant concentration [S]₀ is the initial reactant concentration, k is the rate constant, and t is time. 

 

If we rearrange the equation:

 

[S]/[S]₀ = e^-kt

 

Take the natural log of both sides:

 

-kt = ln([S]/[S]₀)

 

Let's say that t=t_1/2 which is the half life of the reaction, we would then expect (by definition) that [S]=0.5[S]₀ 

 

Substituting and canceling, we get:

 

-kt_1/2 = ln(0.5)

 

By looking that the data provided, we can easily say that 10hrs is the half life since the sucrose concentration drops by half every 10hrs. 

 

Convert into seconds, substituting into the equation and solving for k gets us:

 

k = ln(0.5)/-(10*60*60) = 1.9*10^-5 s^-1

 

The unit is s^-1 because the natural log is unitless due to the numerator and denominator both having units of concentration. 

Answer 2

 Terry has it right. Thank you correcting Terry.

 

Scott