Chemistry Practice Problems Hess's Law Practice Problems Solution: Given the following dataCa(s) + 2C (graphite) → Ca...

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# Solution: Given the following dataCa(s) + 2C (graphite) → CaC2 (s)                      ΔH = -62.8 kJCa(s) + 1/2 O2 (g) → CaO (s)                              ΔH = -635.5 kJCaO (s) + H2O (l) → Ca(OH)2 (aq)                     ΔH = -653.1 kJC2H2 (g) + 5/2O2 (g) → 2CO2 (g) + H2O (l)      ΔH = -1300. kJC(graphite) + O2 (g) → CO2 (g)                        ΔH = -393.5 kJCalculate ΔH for the reactionCaC2 (s) + 2H2O (l) → Ca(OH)2 (aq) + C2H2 (g)

###### Problem

Given the following data

Ca(s) + 2C (graphite) → CaC(s)                      ΔH = -62.8 kJ

Ca(s) + 1/2 O2 (g) → CaO (s)                              ΔH = -635.5 kJ

CaO (s) + H2O (l) → Ca(OH)(aq)                     ΔH = -653.1 kJ

C2H2 (g) + 5/2O2 (g) → 2CO2 (g) + H2O (l)      ΔH = -1300. kJ

C(graphite) + O2 (g) → CO2 (g)                        ΔH = -393.5 kJ

Calculate ΔH for the reaction

CaC2 (s) + 2H2O (l) → Ca(OH)(aq) + C2H2 (g)

###### Solution

To determine the ΔH for the reaction, use Hess Law. Manipulate the individual reactions such that when you add them you arrive to the required reaction.

Required reaction:  CaC2(s) + 2H2O(l)Ca(OH)2(aq) + C2H2(g)

Reaction 1: Ca(s) + 2 C (graphite) → CaC2(s) (reverse reaction)                              ΔH = -62.8 kJ (reverse sign)

(Reverse the reaction because CaC2 is on the product side in the required reaction while it is in the reactant side in Reaction 2)

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