Question

A first order reaction has a rate constant of 2.303x10-³s-¹ The time required for 40g of this
reactant to reduce to 10 g will be-
[Given that log10 2=0.3010]
(1) 230.3 S
(2) 301 s
(3) 2000 S
(4) 602 s

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Answer 1

Answer:

4) 602

Explanation:

T = 2.303 log10 (Initial amount÷ Amount left)

______________________________

Rate Constant

T = 2.303 log10 (40÷10)

______________ Log 10 4 = 0.6020

2.303 × 10-³

T = 0.6020

_____

10-³

T = 602.0 Seconds

Hope this answers the question.