For a first order reaction, A → P, t1/2 (half life) is 10 days. The time required for (1th/4) conversion of A (in days) is: (ln 2=0.693, ln 3=1.1)

Question

For a first order reaction, A → P, t1/2 (half life) is 10 days. The time required for (1th/4) conversion of A (in days) is: (ln 2=0.693, ln 3=1.1) (A) 5 (B) 3.2 (C) 4.1 (D) 2.5

Tardigrade Admin · Accepted Answer

A arrow P For first ordert1/2=(In 2/k)=10 days For t1/4=(In 43/k)=t ⇒ (In2/2In2-In3)=(10/t) t1/4=((2×0.693-1.1)/0.693)×10=4.1

Q. For a first order reaction, $A \to P, t_{1/2}$ (half life) is 10 days. The time required for $\frac{1 ^{t h}}{4}$ conversion of A (in days) is :
(ln 2=0.693, ln 3=1.1)

5

3.2

4.1

2.5

$A \to P$

For first order $t_{1/2} = \frac{I n 2}{k} = 10$ days

For $t_{1/4} = \frac{I n 43}{k} = t$

$\Rightarrow \frac{I n 2}{2 I n 2 - I n 3} = \frac{10}{t}$

$t_{1/4} = \frac{( 2 \times 0.693 - 1.1 )}{0.693} \times 10 = 4.1$

Q. For a first order reaction, A→P,t1/2​ (half life) is 10 days. The time required for 41th​ conversion of A (in days) is : (ln 2=0.693, ln 3=1.1)

5

3.2

4.1

2.5

A→P For first ordert1/2​=kIn2​=10 days For t1/4​=kIn43​=t ⇒2In2−In3In2​=t10​ t1/4​=0.693(2×0.693−1.1)​×10=4.1

Q. For a first order reaction, $A \to P, t_{1/2}$ (half life) is 10 days. The time required for $\frac{1 ^{t h}}{4}$ conversion of A (in days) is :
(ln 2=0.693, ln 3=1.1)

$A \to P$

For first order $t_{1/2} = \frac{I n 2}{k} = 10$ days

For $t_{1/4} = \frac{I n 43}{k} = t$

$\Rightarrow \frac{I n 2}{2 I n 2 - I n 3} = \frac{10}{t}$

$t_{1/4} = \frac{( 2 \times 0.693 - 1.1 )}{0.693} \times 10 = 4.1$