Question

In an enzyme solution, sucrose undergoes fermentation. If $0.10M$ solution of sucrose is reduced to $0.05M$ in $10$ hours and to $0.025M$ in $20$ hours, what is the rate constant ?

• A.
  $1.1\times {10}^{-5}{s}^{-1}$
• B.
  $1.4\times {10}^{-5}{s}^{-1}$
• C.
  $1.9\times {10}^{-5}{s}^{-1}$
• D.
  $0.5\times {10}^{-5}{s}^{-1}$

Answer 1

The correct option is C

$1.9\times {10}^{-5}{s}^{-1}$
Since on doubling the time from 10 hours to 20 hours, fractional reduction of sucrose concentration is also doubled, the reaction must be of the first order.

Since for a first - order reaction, $t_{1/2}=0.693/k_1,hence{k}_1=\frac{0.693}{t_{1/2}}=\frac{0.693}{10\times 60\times 60s}=1.9\times {10}^{-5}{s}^{-1}$