One method of producing ethanol is by the reaction of ethene and water, as represented below.C2H4(𝑔)+H2O(𝑔)⇄C2H5OH(𝑔)Δ𝐻°=−45kJ/mol𝑟𝑥𝑛(g) A chemist wants to run the reaction and maximize the amount of C2H5OH(𝑔) produced. Identify two ways the chemist could change the reaction conditions (other than adding or removing any chemical species) to favor the formation of more product. Justify your answer.

The chemist can manipulate the reaction conditions to favor the formation of more product (C2H5OH) by applying Le Chatelier's principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

1. Temperature: The reaction is exothermic (ΔH° = -45 kJ/mol), meaning it releases heat. According to Le Chatelier's principle, if the temperature is decreased, the reaction will shift to the side that produces heat to counteract this change. Therefore, lowering the temperature will favor the formation of more C2H5OH.

2. Pressure: The reaction has the same number of moles of gas on both sides (1 mole of C2H4 and 1 mole of H2O on the reactants side, and 1 mole of C2H5OH on the products side). Therefore, changing the pressure will not have a significant effect on the position of the equilibrium. However, if we consider the reaction in terms of volume, the reactants side has a larger volume than the product side (because ethene and water are both gases, while ethanol is a liquid at room temperature). Therefore, increasing the pressure would decrease the volume, which would shift the equilibrium to the side with the smaller volume (the product side), thus favoring the formation of more C2H5OH.