Asked by rhea
Calculate the value of ΔHo for the reaction
2 CH4 (g) -> C2H6 (g) + H2 (g)
given the following thermochemical equations:
C2H2 (g) + H2 (g) -> C2H4 (g) ΔHo = – 175.1 kJ
C2H6 (g) -> C2H4 (g) + H2 (g) ΔHo = + 136.4 kJ
C2H2 (g) + 3 H2 (g) -> 2 CH4 (g) ΔHo = – 376.8 kJ
a) + 65.3 kJ
b) + 338.1 kJ
c) – 415.5 kJ
d) + 688.3 kJ
2 CH4 (g) -> C2H6 (g) + H2 (g)
given the following thermochemical equations:
C2H2 (g) + H2 (g) -> C2H4 (g) ΔHo = – 175.1 kJ
C2H6 (g) -> C2H4 (g) + H2 (g) ΔHo = + 136.4 kJ
C2H2 (g) + 3 H2 (g) -> 2 CH4 (g) ΔHo = – 376.8 kJ
a) + 65.3 kJ
b) + 338.1 kJ
c) – 415.5 kJ
d) + 688.3 kJ
Answers
Answered by
DrBob222
Reverse equation 2 and equation 3 and add them to equation 1. See if that isn't the equation you want. When you reverse an equation you change the sign for delta H for that reaction. Then add all of the DH values.
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