A) decreasing the pressure
B) increasing tile temperature
C) removing some \[{{\text{C}}_{\text{2}}}{{\text{H}}_{6}}\]
D) adding some \[{{\text{H}}_{2}}\]
Correct Answer: B
Solution :
Since \[\Delta H\] is negative, therefore, the reaction is exothermic. According to Le-Chatelier's principle, an increase in temperature in this case will favour backward reaction, i.e., will increases the number of moles of\[{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\].You need to login to perform this action.
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