A) 4
B) 8
C) 9
D) 11
Correct Answer: C
Solution :
For a first order process,\[kt=ln\frac{{{[A]}_{0}}}{\left[ A \right]}\] where, \[{{[A]}_{0}}\] = initial concentration. | |
[A] = concentration of reactant remaining at time | |
\[k{{t}_{1/8}}=ln\frac{{{\left[ A \right]}_{0}}}{{{\left[ A \right]}_{0}}/8}=ln\,8\] | ?.. (i) |
\[k{{t}_{1/10}}=ln\frac{{{\left[ A \right]}_{0}}}{{{\left[ A \right]}_{0}}/10}=ln\,10\] | ??(ii) |
Therefore, | |
\[\frac{{{t}_{1/8}}}{{{t}_{1/10}}}=\frac{ln\,\,8}{ln\,\,10}=\log 8=\log {{2}^{3}}3\log 2\] | |
\[\frac{{{t}_{1/8}}}{{{t}_{1/10}}}=3\times 0.3=0.9\] | |
\[\frac{{{t}_{1/8}}}{{{t}_{1/10}}}=10=0.9\times 10=0.9\] |
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