In a first order reaction, R → Q, if ‘k’ is the rate constant and initial concentration of the reactant R is 0.5 M, then the half-life is

Question

In a first order reaction, R → Q, if ‘k’ is the rate constant and initial concentration of the reactant R is 0.5 M, then the half-life is (A) (log 2/k) (B) (log 2/k√0.5) (C) (In 2/k) (D) (0.693/0.5 k)

Tardigrade Admin · Accepted Answer

For a first order reaction, k=(2.303/t) log10 (a/a-x); when t=t1 /2, x=0.5/2 ∴ k=(2.303/t1 /2) log10 (0.5/(0.5/ 2)) ⇒ t1 /2=(2.303/k) log10 2=(in 2/k)

Q. In a first order reaction, R → Q, if ‘k’ is the rate constant and initial concentration of the reactant R is 0.5 M, then the half-life is

$\frac{l o g 2}{k}$

$\frac{l o g 2}{k 0.5}$

$\frac{I n 2}{k}$

$\frac{0.693}{0.5 k}$

For a first order reaction,
$k = \frac{2.303}{t} l o g_{10} \frac{a}{a - x}; w h e n t = t_{1/2}, x = 0.5/2$
$∴ k = \frac{2.303}{t _{1/2}} l o g_{10} \frac{0.5}{( 0.5/2 )}$
$\Rightarrow t_{1/2} = \frac{2.303}{k} l o g_{10} 2 = \frac{in 2}{k}$

Q. In a first order reaction, R → Q, if ‘k’ is the rate constant and initial concentration of the reactant R is 0.5 M, then the half-life is

klog2​

k0.5​log2​

kIn2​

0.5k0.693​

For a first order reaction, k=t2.303​log10​a−xa​;whent=t1/2​,x=0.5/2 ∴k=t1/2​2.303​log10​(0.5/2)0.5​ ⇒t1/2​=k2.303​log10​2=kin2​

$\frac{l o g 2}{k}$

$\frac{l o g 2}{k 0.5}$

$\frac{I n 2}{k}$

$\frac{0.693}{0.5 k}$

For a first order reaction,
$k = \frac{2.303}{t} l o g_{10} \frac{a}{a - x}; w h e n t = t_{1/2}, x = 0.5/2$
$∴ k = \frac{2.303}{t _{1/2}} l o g_{10} \frac{0.5}{( 0.5/2 )}$
$\Rightarrow t_{1/2} = \frac{2.303}{k} l o g_{10} 2 = \frac{in 2}{k}$