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Chemical Kinetics and Nuclear Chemistry question

2024 · 8 Apr · Shift 2 · Q11

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Chemical Kinetics and Nuclear Chemistry
2024 · 8 Apr · Shift 2 · Q11

Chemical Kinetics and Nuclear Chemistry question

2024 · 8 Apr · Shift 2 · Q11

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1

For a reaction

A \xrightarrow{\mathrm{K}_1} \mathrm{~B} \xrightarrow{\mathrm{K}_2} \mathrm{C}

If the rate of formation of B is set to be zero then the concentration of B is given by :

A
$(\mathrm{K}_1-\mathrm{K}_2)[\mathrm{A}]$
B
$(\mathrm{K}_1+\mathrm{K}_2)[\mathrm{A}]$
C
$\mathrm{K}_1 \mathrm{K}_2[\mathrm{~A}]$
D
$(\mathrm{K}_1 / \mathrm{K}_2)[\mathrm{A}]$

View written solutionFree

Correct answer: D

For the consecutive reaction

$A \xrightarrow{K_1} B \xrightarrow{K_2} C$

the rate of change of concentration of intermediate $B$ is

$\frac{d[B]}{dt} = K_1[A] - K_2[B]$

because:

$B$ is formed from $A$ at rate $K_1[A]$
$B$ is consumed to form $C$ at rate $K_2[B]$

The question says that the rate of formation of $B$ is set to zero. This means we apply the steady-state condition:

$\frac{d[B]}{dt} = 0$

So,

$K_1[A] - K_2[B] = 0$

Rearranging,

$K_1[A] = K_2[B]$

$[B] = \frac{K_1}{K_2}[A]$

Now compare with the options:

A: $(K_1-K_2)[A]$ ❌
B: $(K_1+K_2)[A]$ ❌
C: $K_1K_2[A]$ ❌
D: $\left(\frac{K_1}{K_2}\right)[A]$ ✅

Hence, the correct answer is Option D.

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