In: Chemistry
Using the following data, calculate the Hrxnfor the reaction: C2H4(g) + H2(g) →C2H6(g) (ch8)
1.H2(g) + . O2(g) H2O(ℓ)H1= –285.8 kJ
2.C2H4(g) + 3 O2(g) 2 H2O(ℓ) + 2 CO2(g)H2= –1411 kJ
3.3 H2O(ℓ) + 2 CO2(g) 7/2 O2(g) + C2H6(g)H3= –(–1560 kJ)
C2H4(g) + H2(g) →C2H6(g) DHrxn = ?
from the given data
C2H4(g) + 3 O2(g) ------> 2 H2O(ℓ) + 2 CO2(g) DH2 = –1411 kJ
3 H2O(ℓ) + 2 CO2(g) ------> 7/2 O2(g) + C2H6(g) DH3 = – ( –1560 kJ)
H2(g) + 1/2 O2(g) ------------> H2O(ℓ) DH1 = –285.8 kJ
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C2H4(g) + H2(g) ----------------- > C2H6(g) DHrxn = DH1 + DH2 + DH3
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DHrxn = (-285.8-1411+1560) = -136.8 kj