Question

In the following first order completing reaction.
A+ Reagent → Product B Reagent → Product
Calculate ratio of K, / K, when 50% of B will
have been reacted and 94% of A has been reacted
in the same time is -
(I) 4.06
(2) 0.246
(3) 2.06
(4) 0.06
i​

Answer 1

Answer:approx

Explanation:

ForA;K1​∗t=ln100−94100​...(1)

ForB;K2​∗t=ln100−50100​...(2)

Eq.(1)and(2)

K2​/K1​​=log100/6  /log100/50

⇒K2/​K1​​=4.058

Answer 2

Given :

Amount of  A reacted = 50%

Amount of  B reacted = 94%

To Find :

The ration of reaction constants of both the reactions

Solution :

• The reaction rate constant of a reaction is

             $K=\frac{2.303}{t} log\frac{1}{[conc]}$

             $K_{1} =\frac{2.303}{t_{1}} log\frac{100}{50}\rightarrow Equation(1)$

             $K_{2} =\frac{2.303}{t_{2}} log\frac{100}{6}\rightarrow Equation(2)$

• By diving equations 1 & 2

             $\frac{K_{1} }{K_{2}}=\frac{t_{2}}{t_{1}} \times \frac{1.2218}{0.3010}$

• As the time required is same for both the reactants to reach their respective concentrations  

             $\frac{K_{1} }{K_{2}}=\frac{1.2218}{0.3010}$

             $\frac{K_{1}}{K_{2}} =4.06$

The ratio K₁/K₂ is equal to the value 4.06.