5. Use the following data to determine the delta H for the conversion of diamond

Question

5. Use the following data to determine the delta H for the conversion of diamond into graphite: C(diamind) (s) + O2(g) = CO2(g) delta H degrees = -395.4 kJ 2CO2(g) = 2 CO(g) + O2(g) delta H degrees = 566.0 kJ 2CO(g) = C(graphite) (s) + CO2 (g) delta H degrees = -172.5 kJ C(diamond) (s) = C(graphite) delta H degrees = ?

Explanation / Answer

two things to think about.. (1).. multiplying and inverting the equations so that they add to C-d --> Cg (2).. you multiply the ?H's too.. and if you invert, the sign on ?H changes ******* you have this equations (1).. 1 C-d + 1 O2 ---> 1 CO2... ..... .. .?H = -395.4 (2).. 2 CO2.. ... ....---> 2 CO + 1 O2... ?H = +566.0 (3)...2 CO... ... ... .---> 1 C-g + 1 CO2..?H = -172.5 and you want C-d on the left and C-g on the right.. so (1) and (3) don't need to be inverted... (1).. 1 C-d + 1 O2 ---> 1 CO2... ..... .. .?H = -395.4 (2)... ??? (3)...2 CO... ... ... .---> 1 C-g + 1 CO2..?H = -172.5 and you should also be able to see that you'll need to cancel that CO2 on the right of equation #1 with CO2 on the left in equation #2.. so (2) is also in the right direction (1).. 1 C-d + 1 O2 ---> 1 CO2... ..... .. .?H = -395.4 (2).. 2 CO2.. ... ....---> 2 CO + 1 O2... ?H = +566.0 (3)...2 CO... ... ... .---> 1 C-g + 1 CO2..?H = -172.5 but you have 1 CO2 on the right in equation (1) and 1 on the right in equation (3) and you have 2 on the left in equation (2).. so it' looks like the equations simply add up and the CO2's, the CO's and the O2's cancel out. (1).. 1 C-d + 1 O2 ---> 1 CO2... ..... .. .?H = -395.4 (2).. 2 CO2.. ... ....---> 2 CO + 1 O2... ?H = +566.0 (3)...2 CO... ... ... .---> 1 C-g + 1 CO2..?H = -172.5 +___ ____ _____ ____ _____ _____ _____ ____ __ .. .. 1 C-d ---> 1 C-g.. ... .. ... ... ... ... .?H = -395.4 + 566.0 - 172.5 = -1.9

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