Question

Plz Solve This ques​

Answer 1

Answer:

• The 3/4th life of the reaction is 5.4 × 10² sec

Given:

1. Rate constant (K) = 2.54 × 10⁻³ sec⁻¹

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\bigstar\;\underline{\boxed{\sf t=\dfrac{2.303}{k}\;\log\Bigg[\dfrac{a}{a-x}\Bigg]}}$

Here,

• a Denotes initial concentration.
• k Denotes rate constant.
• x Denotes consumed reactants.

Now,

$\bullet\qquad \textsf{For}\;\sf t=t_{3/4}\;,\;x=\dfrac{3\;a}{4}$

Substituting the values,

$\longrightarrow\sf t_{3/4}=\dfrac{2.303}{2.54\times 10^{-3}}\times \log\left[\dfrac{a}{a-\dfrac{3\;a}{4}}\right]\\\\\\\\\longrightarrow\sf t_{3/4}=\dfrac{2.303}{2.54\times 10^{-3}}\times \log\left[\dfrac{a}{\dfrac{4a-3a}{4}}\right]\\\\\\\\\longrightarrow\sf t_{3/4}=\dfrac{2.303}{2.54\times 10^{-3}}\times \log\left[\dfrac{a}{a/4}\right]\\\\\\\\\longrightarrow\sf t_{3/4}=\dfrac{2.303}{2.54\times 10^{-3}}\times \log\left[\dfrac{4}{1}\right]$

$\longrightarrow\sf t_{3/4}=\dfrac{2.303\times 10^{3}}{2.54}\times \log(4)\\\\\\\\\longrightarrow\sf t_{3/4}=\dfrac{2.303\times 10^{3}}{2.54}\times 0.6\\\\\\\\\longrightarrow\sf t_{3/4}=\dfrac{1.38\times 10^{3}}{2.54}\\\\\\\\\longrightarrow\sf t_{3/4}=0.54\times 10^{3}\\\\\\\\\longrightarrow\sf t_{3/4}=5.4\times 10^{2}\\\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf t_{3/4}=5.4\times 10^{2}\;sec}}}}$

∴ The 3/4th life of the reaction is 5.4 × 10² sec.

$\rule{300}{1.5}$

Answer 2

Explanation:

The 3/4th life of the reaction is 5.4 × 10² sec

HOPE this helps you ☺️