Answer to Question #162399 in General Chemistry for Milena Hidalgo

Question

For the reaction:
A⟶products
the following data were obtained:


time / s [A] / M
0.0 1.00
 25  0.80
50   0.67
75 0.57
100 0.50
150 0.40
200 0.33
250 0.29

1) Determine the reaction order 
(2) Determine the speed constant (3 decimal places) 

(3) What is the half-life of this reaction (at the initial concentration)? (1 decimal place) 
 s.

(4) How long will it take for 90% of A to be converted into products? (1 decimal place) 

Answer 1

Q162399

 

For the reaction:
A⟶products
the following data were obtained:


time / s [A] / M
0.0 1.00
 25  0.80
50   0.67
75 0.57
100 0.50
150 0.40
200 0.33
250 0.29

1) Determine the reaction order 
(2) Determine the speed constant (3 decimal places) 

(3) What is the half-life of this reaction (at the initial concentration)? (1 decimal place) 
 s.

(4) How long will it take for 90% of A to be converted into products? (1 decimal place) 

Solution:

1) Determine the reaction order.

Answer :

We are given the reaction A --> products.

Only one reactant is involved in this reaction, so this reaction is 1st order reaction.

2) Detemine the speed constant ( 3 decimal places)

Answer :

The integrated first-order rate equation is

$ln[A_t] -ln[A_0] = -kt ;$

where A0 is the initial concentration of the reactant.

At is the concentration at a time 't'.

k is is the rate constant ( speed constant).

't' is the time.

The above equation can also be written as

$ln[A_t] = -kt + ln[A_0];$ (Equation 1 )

compare this with the straight-line equation, y = m x + c.

where m is the slope. 

plot a graph of ln[At]  versus  time 't' .  This will give you a straight line. 
Below is the graph of ln[At]  verusus time 't'.  

From the graph we have

f(x) = -0.00492989 x - 0.0897

So slope, m = -0.004930s^-1 .

compare it with Equation 1, we have

k = 0.004930 s^-1 .

which in 3 decimal places can be written as k = 0.005 s^-1.

(3) What is the half-life of this reaction (at the initial concentration)? (1 decimal place) .

Answer :

The half-life of the first-order reaction is independent of the concentration.

The half-life equation for the first-order reaction is written as

t_1/2 $= \frac{ln(2)}{k} = \frac{0.693}{k}$

t_1/2 = $= \frac{ln2}{0.004930s^{-1}} = 140.6 s$

Hence the half-life of the given reaction is 140.6 s.

(4) How long will it take for 90% of A to be converted into products? (1 decimal place

Answer :

If we consider that initially there is 100 % of A.

Then after consumption of 90% of A, only 10 % of A will remain.

So [Ao] = 100 % , [At] = 10 % ,

k = 0.004930 s^-1 , t = unknown.

plug all this information in the integrated first-order reaction we have

$ln[A_t] -ln[A_0] = -kt ;$

ln(10) - ln (100) = - 0.004930 s^-1 * t ;

2.3026 - 4.6052 = - 0.004930 s^-1 * t ;

-2.3026 = - 0.004930 s-1 * t ;

divide both the side by -0.004930 s-1 we have

t = $\frac{-2.3026}{-0.004930} = 467.1seconds.$

So it will take 467.1 seconds for 90 % of A to convert to products.