Question

From the following data, C(graphite) + O2(g) → CO2(g) ΔH°rxn = −393.5 kJ/mol H2(g) + 1/2 O2(g) → H2O(l) ΔH°rxn = −285.8 kJ/mol 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l) ΔH°rxn = −3119.6 kJ/mol calculate the enthalpy change for the reaction below: 2 C(graphite) + 3H2(g) → C2H6(g) kJ

Name: enter your answer in the provided box from the following data cgraphite o2g co2gh o rxn 3935 kjmol h2g 1 2 o2g h2olh o rxn 2858 kjmol 2c2h6g 7o2g 4co2g 6h2olh o rxn 31196 kjmol calculate the 78387
Uploaded: 2022-10-29T16:21:54-08:00
Duration: 5 min 28 s
Channel: Sri K
Description: enter your answer in the provided box from the following data cgraphite o2g co2gh o rxn 3935 kjmol h2g 1 2 o2g h2olh o rxn 2858 kjmol 2c2h6g 7o2g 4co2g 6h2olh o rxn 31196 kjmol calculate the 78387

          From the following data,
C(graphite) + O2(g) → CO2(g) ΔH°rxn = −393.5 kJ/mol 
H2(g) + 1/2 O2(g) → H2O(l) ΔH°rxn = −285.8 kJ/mol 
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l) ΔH°rxn = −3119.6 kJ/mol 
calculate the enthalpy change for the reaction below: 
2 C(graphite) + 3H2(g) → C2H6(g) kJ

Added by Gloria T.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Sri K

10/29/2022

Step 1

5 kJ/mol Next, we need to find the enthalpy change for the reaction: H2(g) + 1/2 O2(g) → H2O(l) Given: ΔH°rxn = -285.8 kJ/mol Now, we need to find the enthalpy change for the reaction: 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l) Given: ΔH°rxn = -3119.6 kJ/mol We can Show more…

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From the following data, C(graphite) + O2(g) → CO2(g) ΔH°rxn = −393.5 kJ/mol H2(g) + 1/2 O2(g) → H2O(l) ΔH°rxn = −285.8 kJ/mol 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l) ΔH°rxn = −3119.6 kJ/mol calculate the enthalpy change for the reaction below: 2 C(graphite) + 3H2(g) → C2H6(g) kJ