Question

The half-life of a first order reaction is equal to $245$ seconds. Determine the fraction of reactant a remaining in the reaction after $764$ seconds.

• A. $32$%
• B. $0.28$%
• C. $12$%
• D. $50$%
• E. $88$%

Answer 1

Answer: (C)

$12$%

For a first order reaction,

$t_{1/2}=\frac{0.693}{K}\therefore K=\frac{0.693}{t_{1/2}}=\frac{0.693}{245}=2.83\times {10}^{-3}{sec}^{-1}$

Now, $K=\frac{2.303}{t}\log \frac{\left[A_o\right]}{\left[A_t\right]}\therefore \log \frac{\left[A_o\right]}{\left[A_t\right]}=\frac{2.83\times {10}^{-3}\times 764}{2.303}=0.939\therefore \frac{\left[A_o\right]}{\left[A_t\right]}={10}^{0.939}=8.69\therefore \left[A_t\right]=\frac{1}{8.69}\left[A_o\right]\therefore \left[A_t\right]=0.115\left[A_o\right]$

$\therefore$ Fraction of reactant remaining in reaction=$\frac{\left[A_o\right]}{\left[A_t\right]}\times 100\%=11.5\%\approx 12\%$