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Calculate the amount of reactant in percent that remains after 60 minutes involved in first order reaction. ($k=0.02303\text{ minute}^{-1}$)
Calculate the amount of reactant in percent that remains after 60 minutes involved in first order reaction. ($k=0.02303\text{ minute}^{-1}$)
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Logic Tip: You can also use the integrated rate equation directly: $k = \frac{2.303}{t} \log\left(\frac{100}{100-x}\right)$. Substituting $k=0.02303$ and $t=60$ yields $0.02303 = \frac{2.303}{60} \log\left(\frac{100}{N_t}\right) \implies \log\left(\frac{100}{N_t}\right) = 0.6 \implies \frac{100}{N_t} \approx 4 \implies N_t = 25%$. But the half-life method is much faster!
Updated On: Apr 28, 2026
- 25%
- 75%
- 50%
- 12.5%
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The Correct Option is A
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