Question

A certain first-order reaction has a rate constant of 10 ^-3 sec ^-1. How much time will it take for 10 g of it to reduce to half the quantity

Answer 1

Answer:

From the question, we can write down the following information:

Initial amount = 5 g

Final concentration = 3 g

Rate constant = 1.15 10 - 3s -1

= 444.38 s

= 444 s (approx)

Explanation:

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Answer 2

Given:

k = 10⁻³ s⁻¹

[R]₀ = 10 gm

To Find:

Time taken to reduce to half of its quantity, i.e., the half life of the reaction.

Calculation:

- The half life of a 1st order reaction is not dependent upon the initial concentration.

- The half life of a 1st order reaction is given as:

T1/2 = 0.693 / k

⇒ T1/2 = 0.693 / 10⁻³

⇒ T1/2 = 0.693 × 10³

⇒ T1/2 = 693 seconds

or T1/2 = 11.55 minutes

- So, the reaction will take 693 seconds or 11.55 minutes to reduce 10 g of it into half of its quantity.