A first order reaction has a rate constant of 2.303× 10-3 s-1. The time required for 40g of this reactant to reduce to 10 g will be- [Given that log10 2 = 0.3010

Question

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

Tardigrade Admin · Accepted Answer

For a first order reaction; t1/2 = (0.693/K) t1/2 = (0.693/2.303 × 10-3) = 301 s The time required for 40 g of reactant to reduce to 10g t75 % = 2 × t1/2 t75 % = 2 × 301 = 602 s

Q. A first order reaction has a rate constant of $2.303 \times 1 0^{- 3}$ s $^{- 1}$ . The time required for 40g of this reactant to reduce to 10 g will be-
[Given that $l o g_{10}$ 2 = 0.3010

230.3 s

301 s

2000 s

602 s

For a first order reaction; $t_{1/2} = \frac{0.693}{K}$
$t_{1/2} = \frac{0.693}{2.303 \times 1 0 ^{- 3}} = 301 s$
The time required for $40 g$ of reactant to reduce to $10 g$
$t_{75%} = 2 \times t_{1/2}$
$t_{75%} = 2 \times 301 = 602 s$

Q. A first order reaction has a rate constant of 2.303×10−3 s−1. The time required for 40g of this reactant to reduce to 10 g will be- [Given that log10​ 2 = 0.3010