A) \[1.6\times {{10}^{-8}}M\]
B) \[3.2\times {{10}^{-6}}M\]
C) \[3.2\times {{10}^{-4}}M\]
D) \[1.6\times {{10}^{-4}}M\]
Correct Answer: D
Solution :
\[6HCHO\] \[{{C}_{6}}{{H}_{12}}{{O}_{6}}\] forward reaction \[{{C}_{6}}{{H}_{12}}{{O}_{6}}\] \[\overset{{{K}_{1}}}{leftrightarrows}\] \[6HCHO\] backward reaction \[{{K}_{2}}={{\left[ \frac{1}{{{K}_{1}}} \right]}^{{1}/{6}\;}}\]; \[{{K}_{2}}={{\left[ \frac{1}{6\times {{10}^{22}}} \right]}^{{1}/{6}\;}}\] \[{{K}_{2}}=1.6\times {{10}^{-4}}\]M
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