Consider the following case of completing 1 st order reactions After the start of the reaction at t=0 with only A, the [B] is equal to the [C] at all times. The time in which all three concentrations will be equal is given by

Question

Consider the following case of completing 1 st order reactions <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/4e347b42e053bfa16c7196b7fe7ff6bf-.png" /> After the start of the reaction at t=0 with only A, the [B] is equal to the [C] at all times. The time in which all three concentrations will be equal is given by (A) t=(1/3 k1) ln 2 (B) t=(2/2 k1) ln 3 (C) t=(1/3 k2) ln 2 (D) t=(1/2 k2) ln 3

Tardigrade Admin · Accepted Answer

k1=k2=(2/3) rd of A has reacted for [A]=[B]= [C] ∴ k1+k2=(1/t) ln ([A]0/(1)3[A]0) Or t=(1/k1+k2) ln 3=(1/2 k1) ln 3=(1/2 k2) ln 3

Q. Consider the following case of completing $1$ st order reactions

After the start of the reaction at $t = 0$ with only $A$ , the $[B]$ is equal to the $[C]$ at all times. The time in which all three concentrations will be equal is given by

$t = \frac{1}{3 k _{1}} ln 2$

$t = \frac{2}{2 k _{1}} ln 3$

$t = \frac{1}{3 k _{2}} ln 2$

$t = \frac{1}{2 k _{2}} ln 3$

$k_{1} = k_{2} = \frac{2}{3}$ rd of $A$ has reacted for $[A] = [B] =$ [C]
$∴ k_{1} + k_{2} = \frac{1}{t} ln \frac{[ A ] _{0}}{\frac{1}{3} [ A ] _{0}}$
Or $t = \frac{1}{k _{1} + k _{2}} ln 3 = \frac{1}{2 k _{1}} ln 3 = \frac{1}{2 k _{2}} ln 3$

Q. Consider the following case of completing 1 st order reactions After the start of the reaction at t=0 with only A, the [B] is equal to the [C] at all times. The time in which all three concentrations will be equal is given by

t=3k1​1​ln2

t=2k1​2​ln3

t=3k2​1​ln2

t=2k2​1​ln3