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Standard XII

Physics

Physical Quantities

At 27oC tem...

Question

At $27^{o} C$ temperature, time required for $75$ % completion of a first order reaction is $20$ seconds. What will be its rate constant?

0.693 {s e c}^{- 1}

{m o l}^{- 1}

l t

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0.0693 {s e c}^{- 1}

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0.693 {s e c}^{- 1}

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0.0693 {s e c}^{- 1}

{m o l}^{- 1}

l t

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Solution

The correct option is B $0.0693 {s e c}^{- 1}$
The reaction is 75% complete.
Let,
$a = 100$
$a - x = 100 - 75 = 25$
The rate constant k
$k = \frac{2.303}{t} l o g_{10} \frac{a}{a - x}$
$k = \frac{2.303}{20} l o g_{10} \frac{100}{25}$
$k = 0.0693 s e c^{- 1}$

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At 27oC temperature, time required for 75% completion of a first order reaction is 20 seconds. What will be its rate constant?

The correct option is B 0.0693 sec−1The reaction is 75% complete.Let, a=100a−x=100−75=25The rate constant kk=2.303tlog10aa−xk=2.303 20log1010025k=0.0693sec−1

At $27^{o} C$ temperature, time required for $75$ % completion of a first order reaction is $20$ seconds. What will be its rate constant?

The correct option is B $0.0693 {s e c}^{- 1}$
The reaction is 75% complete.
Let,
$a = 100$
$a - x = 100 - 75 = 25$
The rate constant k
$k = \frac{2.303}{t} l o g_{10} \frac{a}{a - x}$
$k = \frac{2.303}{20} l o g_{10} \frac{100}{25}$
$k = 0.0693 s e c^{- 1}$