Question

Diamon 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 delta H rxn for C(diamond) > C(graphite)
with equations from the following list:

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 AHxn for C(diamond) → C(graphite) with equations from the following list: (1) (3) (4) C(diamond) + O2(g) + CO2(g) 2C02(g) → 2CO(g) + O2(g) C(graphite) + O2(g) → CO2(g) 2CO(g) → C(graphite) + CO2(g) AH = -395.4 kJ AH = 566.0 kJ AH = -393.5 kJ AH = 172.5 kJ

Answer 1

kniven that (1 C(diamond ) +0,(32--CO (32. DH--395.4KJ. (2) 2 CO2( → 2 Colg) +02 (9) DH = 566.04J (3) C (greaphite) +0,(8C02(90% AH = -393.510 (4) 200ls, c(greaphite) +(0₂9) R AH = 172.5KJ we have to find out 1 A Hrexo + Altrexa fora Caliamond ) → (greaphite) preoblem is based on the Hess's lawi This

According to this law, if, A- D DHE A DHI, BOHE c IHM D then OHP = AHg+A Hz + 4H3 Now, adding eq?0 & eq? Caliamond ) +0,00) + CO2(8) 900092 greaphite ) +Cofy Celiamond ) + O2(52+2 co gi - 42002(3)+ graphite ) H5= A He +Atit Now add er ① & ear® Cldiamond ) + 0xf5+260 (5) 2662 + grea) 2 coz (9) — 260ls) toz(8) C(diamond) (greaphite) A Hrexa = AHS + SHE - SH, HAHA + oHg. - 375.4K + 1+3.5kg + 566.0Kg A Hero = 343-1Kg