The half-life period of a first order reaction having rate constant k = 0.231 × 10-10 s-1 will be

Question

The half-life period of a first order reaction having rate constant k = 0.231 × 10-10 s-1 will be (A) 32 × 1010 s (B) 2 × 1010 s (C) 3 × 1010 s (D) 2 × 10-10 s

Tardigrade Admin · Accepted Answer

The half-life of a first order reaction is t1 / 2 =(0.693/k) k =0.231 × 10-10 s -1 So, t1 / 2=(0.693/0.231 × 10-10) ⇒ t1 / 2=3 × 1010 s

Q. The half-life period of a first order reaction having rate constant $k = 0.231 \times 1 0^{- 10} s^{- 1}$ will be

$32 \times 1 0^{10} s$

$2 \times 1 0^{10} s$

$3 \times 1 0^{10} s$

$2 \times 1 0^{- 10} s$

$3 \times 1 0^{- 12} s$

The half-life of a first order reaction is

$t_{1/2} = \frac{0.693}{k}$

$k = 0.231 \times 1 0^{- 10} s^{- 1}$

So, $t_{1/2} = \frac{0.693}{0.231 \times 1 0 ^{- 10}}$

$\Rightarrow t_{1/2} = 3 \times 1 0^{10} s$

Q. The half-life period of a first order reaction having rate constant k=0.231×10−10s−1 will be

32×1010s

2×1010s

3×1010s

2×10−10s

3×10−12s

The half-life of a first order reaction is t1/2​=k0.693​ k=0.231×10−10s−1 So, t1/2​=0.231×10−100.693​ ⇒t1/2​=3×1010s