If k 1 = 4 × 1 0 − 3 sec − 1 & k 2 = 6 × 1 0 − 3 sec − 1 then find the half life of A

For parallel first-order reactions: A k 1 B A k 2 C the effective rate constant is: k t o t a l = k 1 + k 2 Given: k 1 = 4 × 1 0 − 3 s − 1 k 2 = 6 × 1 0 − 3 s − 1 So, k t o t a l = 10 × 1 0 − 3 = 1 0 − 2 s − 1 For a first-order reaction, the half-life is: t 1/2 = k 0.693 Substitute k = 1 0 − 2 s − 1 : t 1/2 = 1 0 − 2 0.693 t 1/2 = 69.3 s Therefore, the correct answer is: (C) 69.31 sec

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ChemistryChemical Kineticskinetics of complexreaction Parallel, Reversible, SequentialMedium2 minQB

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Passage / Comprehension

Consider A parallel reaction following 1 order kinetics

where k₁ & k₂are rate constant for Reaction A → B and A → C respectively

If $k_{1} = 4 \times 1 0^{- 3} sec^{- 1} & k_{2} = 6 \times 1 0^{- 3} sec^{- 1}$

then find the half life of A

Options:

Answer:

Solution:

For parallel first-order reactions:

$A k_{1} B$

$A k_{2} C$

the effective rate constant is:

$k_{t o t a l} = k_{1} + k_{2}$

Given:

$k_{1} = 4 \times 1 0^{- 3} s^{- 1}$

$k_{2} = 6 \times 1 0^{- 3} s^{- 1}$

So,

$k_{t o t a l} = 10 \times 1 0^{- 3} = 1 0^{- 2} s^{- 1}$

For a first-order reaction, the half-life is:

$t_{1/2} = \frac{0.693}{k}$

Substitute $k = 1 0^{- 2} s^{- 1}$ :

$t_{1/2} = \frac{0.693}{1 0 ^{- 2}}$

$t_{1/2} = 69.3 s$

Therefore, the correct answer is:

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:kinetics of complexreaction Parallel, Reversible, Sequential

⏱ 2mℹ️ Source: QB

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