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

Question

The rate of a first order reaction is 0.693× 10-2 mol L-1 min -1 and the initial concentration of the reactants is 1 M, t1/2 is equal to: (A) 6.92 minute (B) 100 minutes (C)  0.693× 10-3 minute (D)  0.693× 10-2 minute

Tardigrade Admin · Accepted Answer

For 1st order reaction, textrate=K× [A] ∴ 0.693× 10-2=K× 1 or K=0.693× 10-2 min -1 We know that, t1/2=(0.693/K)=(0.693/0.693× 10-2) = 100 minute

Q. The rate of a first order reaction is $0.693 \times 10^{- 2} m o l L^{- 1} min^{- 1}$ and the initial concentration of the reactants is 1 M, $t_{1/2}$ is equal to:

6.92 minute

100 minutes

$0.693 \times 10^{- 3}$ minute

$0.693 \times 10^{- 2}$ minute

For 1st order reaction, $rate = K \times [A]$ $∴$ $0.693 \times 10^{- 2} = K \times 1$ or $K = 0.693 \times 10^{- 2} min^{- 1}$ We know that, $t_{1/2} = \frac{0.693}{K} = \frac{0.693}{0.693 \times 10 ^{- 2}}$ = 100 minute

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

6.92 minute

100 minutes

0.693×10−3 minute

0.693×10−2 minute

For 1st order reaction, rate=K×[A] ∴ 0.693×10−2=K×1 or K=0.693×10−2min−1 We know that, t1/2​=K0.693​=0.693×10−20.693​ = 100 minute

Q. The rate of a first order reaction is $0.693 \times 10^{- 2} m o l L^{- 1} min^{- 1}$ and the initial concentration of the reactants is 1 M, $t_{1/2}$ is equal to:

$0.693 \times 10^{- 3}$ minute

$0.693 \times 10^{- 2}$ minute

For 1st order reaction, $rate = K \times [A]$ $∴$ $0.693 \times 10^{- 2} = K \times 1$ or $K = 0.693 \times 10^{- 2} min^{- 1}$ We know that, $t_{1/2} = \frac{0.693}{K} = \frac{0.693}{0.693 \times 10 ^{- 2}}$ = 100 minute