Question: For the reaction \[A \to products\] , \[\dfrac{{ - d\left[ A \right]}}{{dt}} = k\] and at different ...

Question

For the reaction $A \to products$ , $\dfrac{{ - d\left[ A \right]}}{{dt}} = k$ and at different time intervals, [A] values are

Time	$0$	$5\min$	$10\min$	$15\min$
$\left[ A \right]$	$20mol$	$18mol$	$16mol$	$14mol$

At $20$ min, the rate would be:
A. $12$ mol/min
B. $10$ mol/min
C. $8$ mol/min
D. $0.4$ mol/min

Answer 1

Solution

It depends on the concentration.
We need to know that the rate of the reaction is the speed at which a chemical reaction happens. It is expressed with the help of either concentration (amount per unit volume) of a product that is formed in a given time or concentration of a reactant that is consumed in a given unit of time.

Complete step by step answer:
Given,
$A \to products$ , $\dfrac{{ - d\left[ A \right]}}{{dt}} = k$
at $0$ mins it is $20$ mol
at $5$ mins it is $18$ mol
at $10$ mins it is $16$ mol
at $15$ mins it is $14$ mol
For every five minutes interval, the concentration changes by two moles per litre.
Hence, the rate of reaction is constant throughout. We can see that when it is five minutes the concentration becomes eighteen mol after another five mins it is sixteen mol. After fifteen mins it is fourteen mol. So after looking at this we can conclude that after every five minute interval, the concentration changes by two moles per litre.
Rate=Time interval/Change in concentration
$\dfrac{2}{5} = 0.4mol/min$
So the answer is $0.4mol/\min$ .
Which is option D .
And hence D is the correct option for the given solution.

Note:
As we can see that there are changes that are happening in every five mins, it is easy for us to say that there is a proper proportion at which the changes are occurring. The concentration also decreases by two mol. All one needs to do to solve a question is to read the information given and take the necessary information so that they get the correct answer for the question.