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Determine ΔH_rxn for: C (diamond) ---> C (graphite)   With equatios from the...

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Determine ΔH_rxn for:

C (diamond) ---> C (graphite)

 

With equatios from the following list:

 

C (diamond) + O_2 (g) ---> CO_2 (g)       ΔH = -395.4 kJ

2CO_2 (g) ---> 2CO (g)+ O_2 (g)            ΔH = 566.0 kJ 

C (graphite) + O_2 (g) ---> CO_2 (g)       ΔH = -393.5 kJ 

2CO (g) ---> C (graphite) + CO_2 (g)       ΔH = -172.5 kJ

 

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Posted (Answer #1)

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Let us mark the four reactions as:

C (diamond) + O_2 (g) ---> CO_2 (g)       ΔH = -395.4 kJ --- (i)

2CO_2 (g) ---> 2CO (g)+ O_2 (g)            ΔH = 566.0 kJ --- (ii)

C (graphite) + O_2 (g) ---> CO_2 (g)       ΔH = -393.5 kJ --- (iii)

2CO (g) ---> C (graphite) + CO_2 (g)       ΔH = -172.5 kJ--- (iv)

According to Hess’ law of constant heat summation, heat of reaction will be the same so far as the starting and ending compounds are identical, irrespective of the path taken.

Doing a simple mathematical operation, (i) + (ii) + (iv), we get     

C (diamond) + O_2 (g) + 2CO_2 (g) + 2CO (g) ---> CO_2 (g) + 2CO (g)+ O_2 (g) + C (graphite) + CO_2 (g)

Cancelling similar terms on both sides, C (diamond) ---> C (graphite)

This is the required reaction. Hence similar operation involving the heat of reaction values would give the heat of reaction for this reaction.

ΔH = -395.4 + 566.0 + (-172.5) kJ/mol = -1.9 kJ/mol.

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