Carbon occurs in two distinct forms. it can be the soft, black material found in pencils and lock lubricants, called graphite, or it can be the hard, brilliant gem we know as diamond. calculate ∆h0 for the conversion of graphite to diamond for the reaction cgraphite (s) - → cdiamond (s) the combustion reactions are cgraphite (s) + o2 (g) - → co2 (g) ∆h 0 c = - 392.7 kj/mol cdiamond (s) + o2 (g) - → co2 (g) ∆h0 c = - 395.2 kj/mol answer in units of kj.

Question

Kemp

Chemistry

13 September, 13:49

Carbon occurs in two distinct forms. it can be the soft, black material found in pencils and lock lubricants, called graphite, or it can be the hard, brilliant gem we know as diamond. calculate ∆h0 for the conversion of graphite to diamond for the reaction cgraphite (s) - → cdiamond (s) the combustion reactions are cgraphite (s) + o2 (g) - → co2 (g) ∆h 0 c = - 392.7 kj/mol cdiamond (s) + o2 (g) - → co2 (g) ∆h0 c = - 395.2 kj/mol answer in units of kj.

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Babs · Answer 1

Hess' Law can be applied in this case. We will "add" the balanced reactions to produce the required final reaction which is as follows: cgraphite (s) - → cdiamond (s)

Solution:

cgraphite (s) + o2 (g) - → co2 (g) ∆h 0 c = - 392.7 kj/mol (Equation 1)

cdiamond (s) + o2 (g) - → co2 (g) ∆h0 c = - 395.2 kj/mol (Equation 2)

We "flip" equation 2 (so that CO2 and O2 will cancel out) and the value of ∆h0 c will also become positive from negative. Adding up, a value of ∆h0 (rxn) = 2.5 kJ.