5.6 Narrated Notes

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5.6 – Hess’s Law : 

5.6 – Hess’s Law

Hess’s Law : 

Hess’s law states that if a reaction is carried out in a series of steps, ΔH for the overall reaction will equal the sum of the enthalpy changes for the individual steps 5.6 Hess’s Law

Hess’s Law : 

Why is it important that ΔH is a state function? In other words, why can we use Hess’s Law to calculate the ΔH of a reaction? 5.6 Hess’s Law

Hess’s Law : 

The overall enthalpy change for the process in independent of the number of steps or the particular nature of the path by which the reaction occurred It does not matter how we arrived at the target equation 5.6 Hess’s Law

Practice Exercise : 

Carbon occurs in two forms, graphite and diamond. The enthalpy of the combustion of graphite is -393.5 kJ/mol and that of diamond is -395.4 kJ/mol. C(graphite) + O2(g) CO2(g) ΔH = -393.5 kJ C(diamond) + O2(g) CO2(g) ΔH = -395.4 kJ Calculate ΔH for the conversion of graphite to diamond: C(graphite)C(diamond) ΔH = ? 5.6 Practice Exercise

Practice Exercise : 

Calculate ΔH for the conversion of graphite to diamond Use the first reaction as is ∆H =-393.5 kJ You need to flip the second reaction ∆H = +395.4 kJ Add both ∆H’s and ∆H of the reaction is1.9 kJ 5.6 Practice Exercise

Practice Exercise : 

Calculate ΔH for the reaction NO(g) + O(g) NO2(g) given the following information: NO(g) + O3(g) NO2(g) + O2(g)ΔH = -198.9 kJ O3(g) 3/2 O2(g) ΔH = -142.3 kJ O2(g) 2 O(g)ΔH = 495.0 kJ 5.6 Practice Exercise

Practice Exercise : 

Calculate ΔH for the reaction NO(g) + O(g) NO2(g) Use the first reaction as is ∆H = -198.9 kJ Flip the second reaction so ∆H = +142.3 kJ Flip and halve the third reaction so ∆H = -247.5 Add to get ∆Hrxn = -304.1 kJ 5.6 Practice Exercise