The rate for a first order reaction is 0.693 × 10-2 mol L-1 min-1 and the initial concentration of the reactants is 1M, t1/2 is equal to

Question

The rate for a first order reaction is 0.693 × 10-2 mol L-1 min-1 and the initial concentration of the reactants is 1M, t1/2 is equal to (A) 6.932 min (B) 100 min (C) 0.6932 × 10-3 min (D) 0.6932 × 10-2 min

Tardigrade Admin · Accepted Answer

r=k[reactant]-1 ∴ k=(0.693×10-2/1) also t12=(0.693/k)=(0.693/0.693×10-2)=100 min

Q. The rate for a first order reaction is $0.693 \times 1 0^{- 2} m o l L^{- 1} mi n^{- 1}$ and the initial concentration of the reactants is $1 M, t_{1/2}$ is equal to

$6.932$ min

$100$ min

$0.6932 \times 1 0^{- 3}$ min

$0.6932 \times 1 0^{- 2}$ min

$r = k [re a c t an t]^{- 1}$
$∴ k = \frac{0.693 \times 1 0 ^{- 2}}{1}$
also $t_{12} = \frac{0.693}{k} = \frac{0.693}{0.693 \times 1 0 ^{- 2}} = 100 min$

Q. The rate for a first order reaction is 0.693×10−2molL−1min−1 and the initial concentration of the reactants is 1M,t1/2​ is equal to

6.932 min

100 min

0.6932×10−3 min

0.6932×10−2 min