For a first order reaction A®B, the rate constant, k = 5.5 × 10^-¹⁴s^-¹. The time required for 67% completion of reaction is x × 10^-¹ times the half life of reaction. The value of x is ________ (Nearest integer)

(Given : log 3 = 0.4771)

2 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Rate constant, k = 5.5 × 10^-14 s^-1.

$t = \frac{2.303}{k} l o g \frac{{[R]}_{0}}{[R]}$

= $\frac{2.303}{k} l o g 3 - (i)$

$t_{50 %} = \frac{2.303}{k} l o g 2 - (i i)$

From (i) & (ii)

$\frac{t_{67 %}}{t_{50 %}} = \frac{l o g 3}{l o g 2}$

= 1.58 t_50%

So; t_67%is 15.8 × 10^-1 times half life.

X = 16 (the nearest integer)

Rate constant, k = 5.5 × 10-14 s-1. <math> <mrow> <mi>t</mi> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mo>.</mo> <mn>3</mn> <mn>0</mn> <mn>3</mn> </mrow> <mrow> <mi>k</mi> </mrow> </mfrac> <mi>l</mi> <mi>o</mi> <mi>g</mi> <mfrac> <mrow> <msub> <mrow> <mrow> <mo> [</mo> <mrow> <mi>R</mi> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <mrow> <mo> [</mo> <mrow> <mi>R</mi> </mrow> <mo>]</mo> </mrow> </mrow> </mfrac> </mrow> </math>   = <math> <mrow> <mfrac> <mrow> <mn>2</mn> <mo>.</mo> <mn>3</mn> <mn>0</mn> <mn>3</mn> </mrow> <mrow> <mi>k</mi> </mrow> </mfrac> <mi>l</mi> <mi>o</mi> <mi>g</mi> <mn>3</mn> <mtext> </mtext> <mo>−</mo> <mo stretchy="false"> (</mo> <mi>i</mi> <mo stretchy="false">)</mo> </mrow> </math>   <math> <mrow> <msub> <mrow> <mi>t</mi> </mrow> <mrow> <mn>5</mn> <mn>0</mn> <mn>%</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mo>.</mo> <mn>3</mn> <mn>0</mn> <mn>3</mn> </mrow> <mrow> <mi>k</mi> </mrow> </mfrac> <mi>l</mi> <mi>o</mi> <mi>g</mi> <mn>2</mn> <mtext> </mtext> <mo>−</mo> <mo stretchy="false"> (</mo> <mi>i</mi> <mi>i</mi> <mo stretchy="false">)</mo> </mrow> </math>   From (i) & (ii) <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>t</mi> </mrow> <mrow> <mn>6</mn> <mn>7</mn> <mn>%</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>t</mi> </mrow> <mrow> <mn>5</mn> <mn>0</mn> <mn>%</mn> </mrow> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>l</mi> <mi>o</mi> <mi>g</mi> <mn>3</mn> </mrow> <mrow> <mi>l</mi> <mi>o</mi> <mi>g</mi> <mn>2</mn> </mrow> </mfrac> </mrow> </math>  = 1.58 t50%So;         t67% is 15.8 × 10-1 times half life.X = 16 (the nearest integer)

0 Upvotes 0 Downvotes

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

For a first order reaction A®B, the rate constant, k = 5.5 × 10^-¹⁴s^-¹. The time required for 67% completion of reaction is x × 10^-¹ times the half life of reaction. The value of x is ________ (Nearest integer)

(Given : log 3 = 0.4771)

1 Answer

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

For a first order reaction A®B, the rate constant, k = 5.5 × 10-14s-1. The time required for 67% completion of reaction is x × 10-1 times the half life of reaction. The value of x is ________ (Nearest integer) (Given : log 3 = 0.4771)

1 Answer

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

For a first order reaction A®B, the rate constant, k = 5.5 × 10^-¹⁴s^-¹. The time required for 67% completion of reaction is x × 10^-¹ times the half life of reaction. The value of x is ________ (Nearest integer)

(Given : log 3 = 0.4771)