Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
For the reaction \[{{C}_{6}}{{H}_{4}}(g)+{{H}_{2}}{{C}_{2}}{{H}_{6}};\]\[\Delta H=-32.7\,k\,cal\] The equilibrium concentration of\[{{C}_{2}}{{H}_{4}}\]can be increased by (1) decreasing the temperature (2) decreasing the pressure (3) removing some \[{{C}_{2}}{{H}_{6}}\] (4) removing some\[{{H}_{2}}\]A) 1, 2 and 3 are correct.
B) 1 and 2 are correct.
C) 2 and 4 are correct.
D) 1 and 3 are correct.
Correct Answer: C
Solution :
According to Le-Chatelie?s principle, any change in concentration, pressure and temperature shifts the reaction in a direction which undo the effect of change. The reaction, \[{{C}_{2}}{{H}_{4}}(g)+{{H}_{2}}(g){{C}_{2}}{{H}_{6}}(g);\] \[\Delta H=-32.7\text{ }kcal\] Is exothermic, therefore increase in temperature shifts the reaction in backward direction (ie, more\[{{C}_{2}}{{H}_{4}}\]is formed). Number of moles are more at left hand side, therefore decrease in pressure shifts the reaction in backward direction. If some\[{{H}_{2}}\]is removed, then to compensate it, reaction shifts in backward direction
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