A) Increase in concentration of \[+1\]
B) Decrease in concentration of \[+5\]
C) Increase in temperature
D) Increase in total pressure
Correct Answer: C
Solution :
\[{{C}_{6}}{{H}_{5}}C{{H}_{2}}OH\] According to Le-Chateliers principle the conditions favouring formation of I are (i) High concentration of \[C{{H}_{3}}C\equiv CH\] (ii) High temperature (reaction is endothermic so, on increasing the temperature equilibrium will shift in the direction where heat is absorbed). (iii) Low pressure (moles of products are more than moles of reactant, so, if pressure is increased equilibrium will shift in the direction where volume is less).
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