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Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25∘C diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Determine △ Hrxn for C(diamond) → C(graphite) with equations from the following list: (1) C(dianond)+O2(g)⟶CO2(g)ΔH=−395.4kJ (2) 2CO2(g)⟶2CO(g)+O2(g)ΔH=566.0kJ (3) C(graphite)+O2(g)→CO2(g)ΔH=−393.5kJ (4) 2CO(g)⟶C(graphite)+CO2(g)ΔH=−172.5kJ

Answer :

jbferrado

Answer:

-1.9 KJ/mol

Explanation:

In order to solve the problem, we have to rearrange the equations in a way in which all molecules of O₂ and CO₂ are eliminated:

2C(diamond) + 2O₂(g) → 2CO₂(g)     ΔH₁= 2 x (-395.4 KJ) ------> we multiply by 2 both reactants and products

2 CO₂(g) → 2CO(g) + O₂(g)         ΔH₂= 566.0 KJ

CO₂(g) → C(graphite) + O₂(g)     ΔH₃= -1 x (-393.5 KJ) ------> we use reverse rxn

2CO(g) → C(graphite) + CO₂(g)   ΔH₄= -172.5 KJ

When we cancel the molecules that appear both in reactants and products, the total reaction is the following:

2C(diamond) → 2C(graphite)

ΔHt= ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ = 2 x (-395.4 KJ) + 566.0 KJ + (-1 x (-393.5 KJ)) - 172.5 KJ

ΔHt= 347.2 KJ

This is for 2 mol of C(diamond) which are converted in 2 mol of C(graphite). To obtain ΔH for the reaction of 1 mol C(diamond) to 1 mol (graphite) we have to divide into 2:

ΔH= -3.8 KJ/2mol= -1.9 KJ/mol

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