A flask is filled with equal moles of A and B. The half lives of A and B are 100s and 50s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ______________s.

(Given : In 2 = 0.693)

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alok kumar singh | Contributor-Level 10

a month ago

Process is based upon simultaneous disintegration hence,

$\frac{0.693}{100} \times t = 2.303 l o g_{10} \frac{A_{0}}{A_{t}}$ ………….(i)

and $\frac{0.693}{50} \times t = 2.303 l o g_{10} \frac{B_{0}}{B_{t}}$ ………….(ii)

from equation (i) and (ii)

$0.693 t [\frac{1}{50} - \frac{1}{100}] = [l o g \frac{B_{0}}{B_{t}} - l o g \frac{A_{0}}{A_{t}}] \times 2.303$

Here; A₀ = B₀ and $A_{t} = 4 \times B_{t}$

Therefore $0.693 t [\frac{1}{100}] = 2.303 [l o g (\frac{B_{0}}{B_{t}} \times \frac{A_{t}}{A_{0}})]$

$∴ t = \frac{2.303 \times 0.3010 \times 2 \times 100}{693} = 200 s$

Process is based upon simultaneous disintegration hence, <math> <mrow> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>6</mn> <mn>9</mn> <mn>3</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> </mrow> </mfrac> <mo>×</mo> <mi>t</mi> <mo>=</mo> <mn>2</mn> <mo>.</mo> <mn>3</mn> <mn>0</mn> <mn>3</mn> <mi>l</mi> <mi>o</mi> <msub> <mrow> <mi>g</mi> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msub> <mfrac> <mrow> <msub> <mrow> <mi>A</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>A</mi> </mrow> <mrow> <mi>t</mi> </mrow> </msub> </mrow> </mfrac> </mrow> </math> ………….(i)and       <math> <mrow> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>6</mn> <mn>9</mn> <mn>3</mn> </mrow> <mrow> <mn>5</mn> <mn>0</mn> </mrow> </mfrac> <mo>×</mo> <mi>t</mi> <mo>=</mo> <mn>2</mn> <mo>.</mo> <mn>3</mn> <mn>0</mn> <mn>3</mn> <mi>l</mi> <mi>o</mi> <msub> <mrow> <mi>g</mi> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msub> <mfrac> <mrow> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mi>t</mi> </mrow> </msub> </mrow> </mfrac> </mrow> </math>              ………….(ii)from equation (i) and (ii) <math> <mrow> <mn>0</mn> <mo>.</mo> <mn>6</mn> <mn>9</mn> <mn>3</mn> <mi>t</mi> <mrow> <mo>[</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>5</mn> <mn>0</mn> </mrow> </mfrac> <mo>−</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> </mrow> </mfrac> </mrow> <mo>]</mo> </mrow> <mo>=</mo> <mrow> <mo>[</mo> <mrow> <mi>l</mi> <mi>o</mi> <mi>g</mi> <mfrac> <mrow> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mi>t</mi> </mrow> </msub> </mrow> </mfrac> <mo>−</mo> <mi>l</mi> <mi>o</mi> <mi>g</mi> <mfrac> <mrow> <msub> <mrow> <mi>A</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>A</mi> </mrow> <mrow> <mi>t</mi> </mrow> </msub> </mrow> </mfrac> </mrow> <mo>]</mo> </mrow> <mo>×</mo> <mn>2</mn> <mo>.</mo> <mn>3</mn> <mn>0</mn> <mn>3</mn> </mrow> </math>  Here; A0 = B0 and <math> <mrow> <msub> <mrow> <mi>A</mi> </mrow> <mrow> <mi>t</mi> </mrow> </msub> <mo>=</mo> <mn>4</mn> <mo>×</mo> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mi>t</mi> </mrow> </msub> </mrow> </math>  Therefore  <math> <mrow> <mn>0</mn> <mo>.</mo> <mn>6</mn> <mn>9</mn> <mn>3</mn> <mi>t</mi> <mrow> <mo>[</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> </mrow> </mfrac> </mrow> <mo>]</mo> </mrow> <mo>=</mo> <mn>2</mn> <mo>.</mo> <mn>3</mn> <mn>0</mn> <mn>3</mn> <mrow> <mo>[</mo> <mrow> <mi>l</mi> <mi>o</mi> <mi>g</mi> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mi>t</mi> </mrow> </msub> </mrow> </mfrac> <mo>×</mo> <mfrac> <mrow> <msub> <mrow> <mi>A</mi> </mrow> <mrow> <mi>t</mi> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>A</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mo>]</mo> </mrow> </mrow> </math>   <math> <mrow> <mo>∴</mo> <mi>t</mi> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mo>.</mo> <mn>3</mn> <mn>0</mn> <mn>3</mn> <mo>×</mo> <mn>0</mn> <mo>.</mo> <mn>3</mn> <mn>0</mn> <mn>1</mn> <mn>0</mn> <mo>×</mo> <mn>2</mn> <mo>×</mo> <mn>1</mn> <mn>0</mn> <mn>0</mn> </mrow> <mrow> <mn>6</mn> <mn>9</mn> <mn>3</mn> </mrow> </mfrac> <mo>=</mo> <mn>2</mn> <mn>0</mn> <mn>0</mn> <mi>s</mi> </mrow> </math>

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A flask is filled with equal moles of A and B. The half lives of A and B are 100s and 50s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ______________s.

(Given : In 2 = 0.693)

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A flask is filled with equal moles of A and B. The half lives of A and B are 100s and 50s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ______________s. (Given : In 2 = 0.693)

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A flask is filled with equal moles of A and B. The half lives of A and B are 100s and 50s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ______________s.

(Given : In 2 = 0.693)