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COMEDK · Chemistry · 19. Chemical Kinetics

The rate constant for a First order reaction at \(560 \mathrm{~K}\) is \(1.5 \times 10^{-6}\) per second. If the reaction is allowed to take place for 20 hours, what percentage of the initial concentration would have converted to products?

  1. A 10.23
  2. B 21.2
  3. C 11.14
  4. D 12.46
Verified Solution

Answer & Solution

Correct Answer

(A) 10.23

Step-by-step Solution

Detailed explanation

For a first order reaction, the integrated rate equation is given by \(k = \dfrac{2.303}{t} \log \left( \dfrac{[A]_0}{[A]_t} \right)\). Given values are \(k = 1.5 \times 10^{-6} \text{ s}^{-1}\) and \(t = 20 \text{ hours} = 20 \times 3600 \text{ s} = 72000 \text{ s}\).…