JEE MAIN - Chemistry (2019 - 8th April Evening Slot - No. 11)
For a reaction scheme $$A\buildrel {{k_1}} \over
\longrightarrow B\buildrel {{k_2}} \over
\longrightarrow C$$,
if the rate of formation of B is set to be zero then the concentration of B is given by :
if the rate of formation of B is set to be zero then the concentration of B is given by :
$${k_1}{k_2}[A]$$
$$\left( {{{{k_1}} \over {{k_2}}}} \right)[A]$$
$$({k_1} + {k_2})[A]$$
$$({k_1} - {k_2})[A]$$
Explanation
Given,
the rate of formation of B is set to be zero.
$$ \therefore $$ $${{d\left[ B \right]} \over {dt}} = 0$$
For this reaction,
$$A\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{{R_1}}^{{k_1}}} B$$
Rate of reaction for this reaction (R1) = $${k_1}\left[ A \right]$$
For this reaction,
$$B\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{{R_2}}^{{k_2}}} C$$
Rate of reaction for this reaction (R2) = $${k_2}\left[ B \right]$$
Net rate of formation of B = $${{d\left[ B \right]} \over {dt}}$$ = R1 - R2
As $${{d\left[ B \right]} \over {dt}} = 0$$
$$ \therefore $$ R1 - R2 = 0
$$ \Rightarrow $$ $${k_1}\left[ A \right]$$ - $${k_2}\left[ B \right]$$ = 0
$$ \Rightarrow $$ $$\left[ B \right] = {{{k_1}} \over {{k_2}}}\left[ A \right]$$
$$ \therefore $$ $${{d\left[ B \right]} \over {dt}} = 0$$
For this reaction,
$$A\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{{R_1}}^{{k_1}}} B$$
Rate of reaction for this reaction (R1) = $${k_1}\left[ A \right]$$
For this reaction,
$$B\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{{R_2}}^{{k_2}}} C$$
Rate of reaction for this reaction (R2) = $${k_2}\left[ B \right]$$
Net rate of formation of B = $${{d\left[ B \right]} \over {dt}}$$ = R1 - R2
As $${{d\left[ B \right]} \over {dt}} = 0$$
$$ \therefore $$ R1 - R2 = 0
$$ \Rightarrow $$ $${k_1}\left[ A \right]$$ - $${k_2}\left[ B \right]$$ = 0
$$ \Rightarrow $$ $$\left[ B \right] = {{{k_1}} \over {{k_2}}}\left[ A \right]$$
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