Question

the rate constant for first order reaction is 80-s .how time will it take to reduce the concentration of the reactant to 1/18th of its initial value.​

Answer 1

$\bigstar$ Answer $\bigstar$

$\checkmark$ Given:

The Rate Constant for First Order Reaction is 80 s⁻¹

$\checkmark$ To Find:

Time taken to reduce the concentration of the reactant to 1/18th of its initial value

$\checkmark$ Solution:

We know that,

$\star$ First Order Reaction

$\orange{\implies \boxed{\rm t=\dfrac{2.303}{k}log\dfrac{[A_{0}]}{[A]}}}$

Here,

• k = Rate Constant
• t = Time
• A₀ = Initial Concentration
• A = Final Concentration

Given that,

The Rate Constant for First Order Reaction is 80 s⁻¹

Therefore,

k = 80 s⁻¹

Given that,

To reduce the concentration of the reactant to 1/18th of its initial value.

Therefore,

$\bold{\rm A_{0}=x}$

Hence,

$\bold{\rm A=\dfrac{x}{18}}$

According to the Question,

We are asked to find the Time taken to reduce the concentration of the reactant to 1/18th of its initial value

Hence,

Substituting the values we get,

We get,

$\blue{\implies \rm t=\dfrac{2.303}{80 \ s^{-1}} \ log \left(\dfrac{x}{\dfrac{x}{18}}\right)}$

$\green{\implies \rm t=\dfrac{2.303}{80 \ s^{-1}} \times log \ 18}$

$\red{\implies \rm t=\dfrac{2.303}{80 \ s^{-1}} \times 1.255}$

$\sf Since,$

$\pink{\sf \longrightarrow log \ 18=1.2552}$

$\gray{\rm \implies t=\dfrac{2.303 \times 1.255}{80} \ s}$

$\blue{\rm \implies t=\dfrac{2.890}{80} \ s}$

$\green{\rm \implies t=0.0361 \ s}$

$\red{\rm \implies t=3.6 \times 10^{-2} \ s}$

Answer 2

Answer:

$\bigstar$ Answer $\bigstar$

$\checkmark$ Given:

The Rate Constant for First Order Reaction is 80 s⁻¹

$\checkmark$ To Find:

Time taken to reduce the concentration of the reactant to 1/18th of its initial value

$\checkmark$ Solution:

We know that,

$\star$ First Order Reaction

$\orange{\implies \boxed{\rm t=\dfrac{2.303}{k}log\dfrac{[A_{0}]}{[A]}}}$

Here,

k = Rate Constant

t = Time

A₀ = Initial Concentration

A = Final Concentration

Given that,

The Rate Constant for First Order Reaction is 80 s⁻¹

Therefore,

k = 80 s⁻¹

Given that,

To reduce the concentration of the reactant to 1/18th of its initial value.

Therefore,

$\bold{\rm A_{0}=x}$

Hence,

$\bold{\rm A=\dfrac{x}{18}}$

According to the Question,

We are asked to find the Time taken to reduce the concentration of the reactant to 1/18th of its initial value

Hence,

Substituting the values we get,

We get,

$\blue{\implies \rm t=\dfrac{2.303}{80 \ s^{-1}} \ log \left(\dfrac{x}{\dfrac{x}{18}}\right)}$

$\green{\implies \rm t=\dfrac{2.303}{80 \ s^{-1}} \times log \ 18}$

$\red{\implies \rm t=\dfrac{2.303}{80 \ s^{-1}} \times 1.255}$

$\sf Since,$

$\pink{\sf \longrightarrow log \ 18=1.2552}$

$\gray{\rm \implies t=\dfrac{2.303 \times 1.255}{80} \ s}$

$\blue{\rm \implies t=\dfrac{2.890}{80} \ s}$

$\green{\rm \implies t=0.0361 \ s}$

$\red{\rm \implies t=3.6 \times 10^{-2} \ s}$