[Chem] 急!! Enthalpy問題!! 20分!!

2014-01-09 4:21 am
1. Given that:
6C(graphite) + 6H2(g) +3O2(g) >> C6H12O6(s) Delta H1 = -1247 kJ mol^-1
C(graphite) + O2(g) >> CO2(g) Delta H2 = -393.5 kJ mol^-1
H2(g) +1/2O2(g) >> H2O(l) Delta H3 = -286.0 kJ mol^-1

a.Construct an enthalpy change cycle to determine the standard enthalpy change of combustion for glucose.

b.Construct an enthalpy level diagram for the combustion of glucose.

2. Given that the standard enthalpy of formation of liquid methanol, carbon dioxide gas and liquid water are -238.5, -393.5 and -285.8 kJ mol^-1.

a.Determine the standard enthalpy change of combustion of liquid methanol.

3.Given that:
C2H2(g) + 2H2(g) >> C2H6(g) Delta H1 = -311.0 kJ mol^-1
H2(g) + 1/2O2(g) >> H2O(l) Delta H2 = -286.0 kJ mol^-1
C2H6(g) + 7/2O2(g) >> 2CO2(g) +3H2O(l) Delta H3 = -1560 kJ mol^-1

a.Construct an enthalpy change cycle to determine the standard enthalpy change of combustion.

4.100cm^3 of 1.0M sodium hydroxide solution and 100 cm^3 of 1.0M ethanoic acid were mixed in a calorimeter. The maximum temperature rise was found to be 5.3 celcius degree. (given that the specfic heat capacity and the density of the reaction mixture are 4.2 J g^-1 K^-1 and 1.0gcm^-3 respectively, the heat capacity of calorimeter is 90.0JK^-1)

a.Calculate the standard enthalpy change of neutralization of sodium hydroxide solution and ethanoic acid.

b.The theoretical value of the standard enthalpy change of neutralization for a strong acid and a strong alkali is -57.3 kJmol^-1. Account for the discrepancy between the theoretical value and the value obtained in (a).

各位幫幫忙!!!!
更新1:

1a. C6H12O6 + 6O2 >> 6CO2 + 6H2O Enthalpy change of combustion for glucose: 6x(-393.5)+6x(-286)-(-1274) = -2806 Is that right? I know how to write the equation but I don't know how to draw cycle and level diagram, can someone help me?

回答 (2)

2014-01-14 3:13 am
✔ 最佳答案
"幫忙" 的意思不是幫你做功課. 請你說明你有何地方不懂, 例如是何時要加何時要減何時要乘除, 或是如何畫cycle, 或是不明白要計甚麼, 等等.
你可以試做一題並打上來, 讓我們幫你看看做得正確與否; 或者可以為你做一題作示範.

2014-01-08 23:09:52 補充:
順便呼籲: 勿為數十分而輕易幫人做功課. 為了一個虛銜而阻人學習, 實在很無謂.

2014-01-09 18:45:20 補充:
哈哈, 多謝賞面! 字數字數字數

2014-01-13 19:13:50 補充:

圖片參考:http://imgcld.yimg.com/8/n/HA00075708/o/20140113185707.jpg


To draw a cycle,
1.
write down the equation that you'd like to find its enthalpy change.

2.
since you have 3 other equations given, you can link them up to form another "path".
for example, apart from the equation concerned, there's one more equation involving C6H12O6 (the arrow on the left).
notice that you still have to balance the equation, e.g. adding the oxygen.

3.
put the enthalpy changes on the arrows. i've inserted some circles to help showing the reactants concerned.
notice that the enthalpy changes also requires multiplying with the coefficient.
for example, H2 + (1/2)O2 ------> H2O , (delta) H1
there're six H2 reacting to give 6 H2O, so multiply the enthalpy change by 6.

4.
overall enthalpy change: if you've learnt a bit of vector or mechanics in physics, it'd be easy to understand that
arrow of (delta)H = reverse of dH1 + 6x dH3 + 6x dH2
= -(dH1) + 6x(dH3) + 6x(dH2)


btw, dH1 is -1247 kJ/mol, not -1274 kJ/mol .





圖片參考:http://imgcld.yimg.com/8/n/HA00075708/o/20140113185711.jpg

for level diagram, first identify the highest-energy and lowest-energy species.
then, say, from the top, draw the species under it.
compare the energy levels between the 2 species in the middle.
you don't have to be proportional, but have to follow the order.




2014-01-15 04:00:00 補充:
呵呵, 對的, 是我懶惰了. 確實要加physical state.
2014-01-10 12:38 am
己式庚辛 :

跟你意見啦!

2014-01-13 21:27:52 補充:
Sorry! 一個意見,沒有了 physical state 有點問題。


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