A) \[104\]
B) \[52153\]
C) \[Alkyne\xrightarrow[Lindlars\,\,catalyst]{{{H}_{2}}}A\xrightarrow{Ozonolysis}\underset{only}{\mathop{B}}\,\]
D) \[\xleftarrow[\Pr ocess]{Wac\ker }C{{H}_{2}}=C{{H}_{2}}\]
Correct Answer: C
Solution :
\[{{N}_{2}}+2{{O}_{2}}\rightleftharpoons 2N{{O}_{2}}\] \[{{K}_{1}}=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}]{{[{{O}_{2}}]}^{2}}}\] or \[100=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}]{{[{{O}_{2}}]}^{2}}}\] ??..(i) Again \[[N{{O}_{2}}]\rightleftharpoons \frac{1}{2}{{N}_{2}}+{{O}_{2}}\] \[{{K}_{2}}=\frac{{{[{{N}_{2}}]}^{1/2}}[{{O}_{2}}]}{[N{{O}_{2}}]}\] or \[K_{2}^{2}=\frac{[{{N}_{2}}]{{[{{O}_{2}}]}^{2}}}{[N{{O}_{2}}]}\] ????(ii) Eqs. (i) x (ii), we get \[100\times K_{2}^{2}=1\] or \[K_{2}^{2}=\frac{1}{100}\] or \[{{K}_{2}}=\frac{1}{10}=0.1\]
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