A) \[{{K}_{1}}\]
B) \[{{K}_{1}}{{I}_{abs}}\]
C) \[\frac{{{K}_{1}}{{K}_{3}}{{I}_{abs}}}{{{K}_{2}}}\]
D) \[\frac{{{K}_{1}}{{I}_{abs}}}{{{K}_{2}}}\]
Correct Answer: B
Solution :
\[\frac{d[{{C}_{24}}{{H}_{28}}]}{dt}=\frac{{{K}_{1}}{{K}_{3}}{{I}_{abs}}[{{C}_{12}}{{H}_{14}}]}{{{K}_{2}}+{{K}_{3}}[{{C}_{12}}{{H}_{14}}]}=\frac{{{K}_{1}}{{K}_{3}}{{[I]}_{absorb}}}{{{K}_{2}}+{{K}_{3}}}={{K}_{1}}{{[I]}_{absorb}}\]\[{{C}_{12}}{{H}_{14}}\frac{{{K}_{1}}{{K}_{3}}{{[I]}_{absorb}}}{{{K}_{2}}+{{K}_{3}}}={{K}_{1}}{{[I]}_{absorb}}{{C}_{12}}{{H}_{14}}*\] \[{{C}_{12}}{{H}_{14}}^{*}+{{C}_{12}}{{H}_{14}}\xrightarrow[{}]{{{K}_{3}}}{{C}_{24}}{{H}_{28}}\]\[[B]\]as\[[{{C}_{2}}{{H}_{14}}]\]in excess.You need to login to perform this action.
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