JEE MAIN - Chemistry (2022 - 24th June Evening Shift - No. 5)
For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is
(Given : ln 10 = 2.303 and log 2 = 0.3010)
1.12
2.43
3.32
33.31
Explanation
$$\mathrm{A} \rightarrow$$ Products
For a first order reaction,
$$ \mathrm{t}_{1 / 2}=\frac{\ln 2}{\mathrm{k}}=\frac{0.693}{\mathrm{k}} $$
Time for $$90 \%$$ conversion,
$$t_{90 \%}=\frac{1}{k} \ln \frac{100}{10}=\frac{\ln 10}{k}=\frac{2.303}{k}$$
$$t_{90\%}=\frac{2.303}{0.693} t_{1 / 2}=3.32 t_{1 / 2}$$
For a first order reaction,
$$ \mathrm{t}_{1 / 2}=\frac{\ln 2}{\mathrm{k}}=\frac{0.693}{\mathrm{k}} $$
Time for $$90 \%$$ conversion,
$$t_{90 \%}=\frac{1}{k} \ln \frac{100}{10}=\frac{\ln 10}{k}=\frac{2.303}{k}$$
$$t_{90\%}=\frac{2.303}{0.693} t_{1 / 2}=3.32 t_{1 / 2}$$
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