Question
Question: When the concentration of a reactant in reaction \(A \rightarrow B\) is increased by \(8\) times but...
When the concentration of a reactant in reaction A→B is increased by 8 times but rate increases only 2 times, the order of the reaction would be:
(A) 2
(B) 31
(C) 4
(D) 21
Solution
This question is based on the concept of rate and order of a reaction. The order of a reaction is the power to which the concentration of a reactant is raised. It is the stoichiometric coefficient of the reactant in a reaction. A reaction involving only a single reactant is called a first-order reaction. In a first-order reaction, the rate is dependent on the concentration of only one reactant with its stoichiometric coefficient raised to its power.
Complete step by step answer:
The rate of a reaction involving a single reactant is given by the following formula:
r=k[R]n
Where r is the rate of the equation;
k is the rate constant or the proportionality constant;
[R] is the concentration of the reactant
n is the order of the reaction.
For the reaction, A→B, the reactant is A. So, the rate equation for the given reaction is as follows:
r=k[A]n …(i)
When the concentration of the reaction is increased by 8 times, it becomes [8A].
When the rate increases two times, the new rate becomes 2r.
Therefore, according to the question, the new rate equation can be represented as:
2r=k[8A]n …(ii)
Dividing equation (ii) by (i), we get;
r2r=k[A]nk[8A]n
⇒2=8n
⇒21=(23)n
⇒21=(2)3n
⇒1=3n
⇒n=31
Therefore, the value of the order of the reaction, n is equal to 31.
So, the correct answer is Option B .
Note: The difference between the rate and rate constant of a reaction is that the rate (r) of a reaction depends upon the concentration of the reactant at a particular temperature, and is variable, whereas, the rate constant (k) of a reaction is constant (fixed) at a particular temperature, and is invariable. According to the rule of exponentiation, the powers (exponents) of two common bases are the same. For example – If ax=ay, then x=y.