For a reaction, A → B + C, it was found that at the end of 10 minutes from the start, the total optical rotation of the system was 50° and when the reaction is complete, it was 100°. Assuming that only B and C are optically active and dextrorotatory. Calculate the rate constant of this first order reaction

Question

For a reaction, A → B + C, it was found that at the end of 10 minutes from the start, the total optical rotation of the system was 50° and when the reaction is complete, it was 100°. Assuming that only B and C are optically active and dextrorotatory. Calculate the rate constant of this first order reaction (A) k = 0.695 min-1 (B) k = 0.0693 sec-1 (C) k = 0.0693 min-1 (D) k = 0.00693 sec-1

Tardigrade Admin · Accepted Answer

r10=50°,r∞=100°,r0=0 ∴ k=(2.303/10)log ((100-0/100-50))=0.0695 min-1

$k = 0.695 mi n^{- 1}$

$k = 0.0693 se c^{- 1}$

$k = 0.0693 mi n^{- 1}$

$k = 0.00693 se c^{- 1}$

$r_{10} = 5 0^{\circ}, r_{\infty} = 10 0^{\circ}, r_{0} = 0$
$∴ k = \frac{2.303}{10} l o g (\frac{100 - 0}{100 - 50}) = 0.0695 mi n^{- 1}$

k=0.695min−1

k=0.0693sec−1

k=0.0693min−1

k=0.00693sec−1

r10​=50∘,r∞​=100∘,r0​=0 ∴k=102.303​log(100−50100−0​)=0.0695min−1

$k = 0.695 mi n^{- 1}$

$k = 0.0693 se c^{- 1}$

$k = 0.0693 mi n^{- 1}$

$k = 0.00693 se c^{- 1}$

$r_{10} = 5 0^{\circ}, r_{\infty} = 10 0^{\circ}, r_{0} = 0$
$∴ k = \frac{2.303}{10} l o g (\frac{100 - 0}{100 - 50}) = 0.0695 mi n^{- 1}$