Hess’s Law The heat evolved or absorbed in a chemical process is the same whether the process takes place in 1 or several steps.

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Hess’s Law The heat evolved or absorbed in a chemical process is the same whether the process takes place in 1 or several steps

Example 1 A man works hard his whole life and retires with a personal wealth of $2 000 000. A man makes and loses a $2 000 000 several times through his life but eventually retires with a personal wealth of $2 000 000. A man lives in poverty until the day he retires and then wins $2 000 000 in a lottery. Each man traveled a different path, but all ended up with 2 million dollars. Their wealth is therefore independent of how their lives were lived.

Example 2 A student walks from MacNab to Westcliffe Mall to meet friend and together they walk to Silver City to see a movie. Another student walks directly from MacNab to Silver City. Both students started in the same spot, and finish in the same spot but travel different "paths" , and distances to get there. Displacement is independent of path taken. It depends only the beginning and end point.

Scientists have determined that enthalpy change is independent of how the system changes from beginning to end.

This suggests two new ideas about chemical reactions. reactions do not occur in a single step but rather by a series of steps. 2) reactions can take various "paths to completion". These paths or steps may be different yet produce the same result. Several reaction pathways may produce the same result.

Example 3: A + B  D + E could take 2 steps 1) A + B  C ΔH = 100 kJ 2) C  D + E ΔH = -200 kJ

Target Equation: Fe2O3 + 2Al  Al2O3 + 2Fe ΔH = ? Given: 1) 2Al + 3/2 O2  Al2O3 ΔH = -1675.7 kJ 2) 2Fe + 3/2 O2  Fe2O3 ΔH = -824.2 kJ

Target Equation: H2O + C  CO + H2 ΔH = ? Given: 1) 2C + O2  2CO ΔH = -221.0 kJ 2) 2H2 + O2  2H2O ΔH = -483.6 kJ

Target Equation: CO + H2 + O2  CO2 + H2O ΔH = ? Given: 2C + O2  2CO ΔH = -221.0 kJ C + O2  CO2 ΔH = -393.5 kJ 3) 2H2 + O2  2H2O ΔH = -483.6 kJ

3H2(g) + CO(g)  CH4 + H2O(g) ΔH = ? Given: 1) 2H2(g) + O2(g)  2H2O(g) ΔH = -484 kJ 2) 2C(s) + O2(g)  2CO(g) ΔH = -221 kJ 3) CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ΔH = -803 kJ 4) C(s) + O2(g)  CO2(g) ΔH = -394 kJ