How rate of reaction and concentration of reactant, affect rate constant

As we discussed in previous that there is a certain relationship between the rate constant, rate of reaction and the concentration of reactant species and this relationship is given by "Rate law". 
According to the Rate Law, "The rate of reaction is directly proportional to the concentration of the reactants species and if we remove that proportionality sign, we add a new constant which is known as proportionality constant or reaction rate constant".

Thus, the rate of reaction is depending on the values of reaction rate constant and the concentration of reactant species. In general, the rate law is expressed as,
                        A ➝ products
   Rate Law of the above reaction is given as,
                    ➩ rate, -r = k [A] ------(1)
Here, '-' sign represent that the concentration of reactant species is continuously decreases over the time. 
Hence;
Rate of Reaction ＝ (Reaction Rate Constant ❌ Concentration of Reactant species )
After rearrange the above equation, we get;
                   ➩ k = -r / [A] 
                                  ------------------(2)
According to the above equation, we can say that the rate constant for a reaction is directly proportional to the rate of reaction while inversely proportional to the concentration of reactant species. Thus, the higher the rate of reaction, the higher the reaction rate constant and the lower the concentration of reactant molecules, the higher the reaction rate constant.
     
Let us look some of the situation in which the changes in reactant's concentration and reaction rate is affect the reaction rate constant for that reaction.

Que- What is rate constant when rate of reaction doubles and the concentration of reactant increases 4 times ??
Solve
As we know that;
            Rate constant ∝ Rate of reaction
      Rate constant ∝ Concentration of reactant

Concentration of reactant species generally represent by "[Reactant]". In a combined form,
➩ Rate constant ∝ Rate/ [Reactant]

Say, A is a reactant whose concentration is represented as [A] , r represent the rate of reaction and k represent the reaction rate constant then,
         ➩ Rate constant, k ∝(or =) r / [A]
                           Or
         ➩ Rate constant, k = r / [A]
                                         --------------(1)
According to the question, if the rate of reaction is getting double and the concentration of reactant increases 4 times then the new rate constant,
        ( ∵ r= 2r and [A]= 4[A] )
    Rate constant, k' = 2r / 4[A]
                  ➩ k' = r / 2[A]
                  ➩ k' = (r/[A]) / 2
                          { ∵ from equation (1); k= r/[A] }
                  ➩ k' = k/2
Thus, the new reaction rate constant will be half of the old rate constant.

QUE- What is rate constant when rate of reaction doubles and the concentration of the reactant is reduced to half ??
Solve- 
As we know that,
       ➩ Rate constant ∝ { Rate / [reactant] }
Say, A is a reactant whose concentration is represented as [A] , r represent the rate of reaction and k represent the reaction rate constant then,
          ➩ Rate constant, k = r / [A]
                                          ---------------(1)
According to the question, 
                 r= 2r and [A]= [A] /2
Then the new rate constant is given as;
       Rate constant, k' =  2r / { [A]/2 }
                  ➩ k' = 4r/[A]
                  ➩ k' = 4 (r/[A])
                          { ∵ from equation (1); k= r/[A] }
                  ➩ k' = 4 k
Thus, the new reaction rate constant will be four times the old rate constant.


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