Posted by John on .
Given the following hypothetical equations,
2A + B → 3C + 2D ΔH = -46.7kJ
C → B ΔH = 13.1kJ
2E → B ΔH = 84.5kJ,
what is the enthalpy change (kJ) for the following reaction?
A → D + 2E
Add eqn 1 as is to 3x eqn 2 to 2x the reverse of eqn 3. That will give you 2x the equn you want so take 1/2 of the final equation. When reversing an equation, change the sign of dH. When multiplying an equation, multiply dH by the same number. Add all of them take 1/2 final equation and 1/2 final dH.