Posted by **John** on Wednesday, March 7, 2012 at 9:21pm.

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

- Chemistry -
**DrBob222**, Wednesday, March 7, 2012 at 9:45pm
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.

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