The standard enthalpy change for the following reaction is -235 kJ at 298 K. 2 C(s,graphite) + 3 H2(g) + 1/2 O2(g) CH3CH2OH(g) ΔH° = -235 kJ

Question

What is the standard enthalpy change for the reaction at 298 K?

1. 2 CH₃CH₂OH(g)  4 C(s,graphite) + 6 H₂(g) + O₂(g

Answer 1

2C(s,graphite) + 3H2(g) + 1/2 O2(g) ==> CH3CH2OH(g) ... ∆Hº = -235 kJ

2 CH3CH2OH(g) ==> 4C(s,graphite) + 6H2(g) + O2(g) is the reverse of the above reaction (change the sign of ∆H) AND it is also twice (x2) the above reaction, so multiply ∆H by 2.

∆H for this reaction = +235 kJ x 2 = +470 kJ