Answer:
-471 Kj/mole acrylic acid
Explanation:
THIS IS THE COMPLETE QUESTION BELOW;
There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In the first step, calcium carbide and water react to form acetylene and calcium hydroxide: CaC (s) + 2 H2O(g) - CH (9) + Ca(OH),(s) AH -414. kJ In the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6 C H (9) + 3 CO2(9) + 4H2O(g) - SCH,CHCO,H) AH-132. kJ Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ?
The two equations from the reaction can be written as;
a)CaC₂(s) + 2H₂O(l) ------->C₂H₂(g) + CaOH₂(s)
Δ H= -414Kj ........................ equation (a)
b)6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj ...................... equation (b)
In equation (b)acrylic acid was produced by the reaction between Acetylene carbon dioxide and water
Then we can multiply equation(a) by factor of 6 and the ΔH Then we have (6× -414Kj)= ΔH= -2484Kj.
6CaC₂(s) + 12H₂O(l) ------->6C₂H₂(g) + 6CaOH₂(s)
Δ H= -2484Kj.................. equation (c)
6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj
Then add equation (c) and equation(b) then we have
6CaC₂(s) + 16H₂O(l)+3CO₂(g)------> 5CH₂CHCO₂H(g) + 6CaOH₂(s) ΔH= -2352Kj
ΔH(net)= -2352Kj/5moles
=-471Kj/mole
therefore, net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ? is -471Kj/mole acrylic acid
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