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 andwater react to form acetylene and calcium hydroxide:CaC2(s) + 2 H20(g)C2H2(g) Ca(OH)2(s)AH=-414. kJIn the second step, acetylene, carbon dioxide and water react to form acrylic acid :6 C2H2(9) + 3 CO2(g) + 4H20(g)5 CH2CHCO2H(g)AH=132. kJCalculate 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.kJ?XSubmit AssignmentContinuePrivacyTerms of Use2019 McGraw-Hill Education. All Rights Reserved.DDDIIO00O00F10F9F8F7escF6F 5F 4F3F2F100

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Asked Nov 18, 2019
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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:
CaC2(s) + 2 H20(g)C2H2(g) Ca(OH)2(s)
AH=-414. kJ
In the second step, acetylene, carbon dioxide and water react to form acrylic acid :
6 C2H2(9) + 3 CO2(g) + 4H20(g)5 CH2CHCO2H(g)
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.
kJ
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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: CaC2(s) + 2 H20(g)C2H2(g) Ca(OH)2(s) AH=-414. kJ In the second step, acetylene, carbon dioxide and water react to form acrylic acid : 6 C2H2(9) + 3 CO2(g) + 4H20(g)5 CH2CHCO2H(g) 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. kJ ? X Submit Assignment Continue Privacy Terms of Use 2019 McGraw-Hill Education. All Rights Reserved. DD DII O00 O00 F10 F9 F8 F7 esc F6 F 5 F 4 F3 F2 F1 00

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Expert Answer

Step 1

The standard enthalpy of a reaction is the change in the enthalpy which occurs in a system when the matter gets transformed by a given chemical reaction, provided that all the reactants and products are present in their standard states. The enthalpy for a given reaction can be calculated as follows:

ΔΗΔΗ
-ΔΗ
products
reacants
π1
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ΔΗΔΗ -ΔΗ products reacants π1

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Step 2

The first reaction equation is as follows:

The second equation is as follows:

СаС,(s)+2H,О(1) —>С,Н, (g)+Са(ОН), (s) ДН-414 kJ
6С Н, (9)+3СО, (g)+4H,О(g)-
5CH,CHCO,H(g) ДН-132 kJ
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СаС,(s)+2H,О(1) —>С,Н, (g)+Са(ОН), (s) ДН-414 kJ 6С Н, (9)+3СО, (g)+4H,О(g)- 5CH,CHCO,H(g) ДН-132 kJ

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Step 3

Since only one mole of ethylene is produce...

{Сас, (s)+2H,О()—сН.(8)+Са(ОН), (6) } х6
(ДН-414 k.) х 6
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{Сас, (s)+2H,О()—сН.(8)+Са(ОН), (6) } х6 (ДН-414 k.) х 6

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