A) 1, 1
B) 1, 2
C) 0, 1
D) 1, 0
Correct Answer: A
Solution :
During any experiment, pH is constant, hence |
\[\frac{-d[A]}{dt}=K'{{[A]}^{a}}\] where \[K'=K{{[{{H}^{+}}]}^{b}}\] |
As \[{{t}_{1/2}}\] is independent of initial conc. So \[a=1\] |
Also K' can be given by \[=\frac{0.693}{{{t}_{1/2}}}\] |
\[\frac{{{({{t}_{1/2}})}_{1}}}{{{({{t}_{1/2}})}_{2}}}=\frac{K_{2}^{'}}{K_{1}^{'}}=\frac{K[{{H}^{+}}]_{2}^{b}}{K[{{H}^{+}}]_{1}^{b}}=\frac{[{{H}^{+}}]_{2}^{b}}{[{{H}^{+}}]_{1}^{b}}\] |
Now \[\frac{100}{10}=\left( \frac{{{10}^{-4}}}{{{10}^{-5}}} \right)\text{or}\,b=1\] |
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