chem 2 question ?

Question

Calculate H for the following reaction,

CaO(s) + CO2(g) --> CaCO3(s)

given the thermochemical equations below.

2 Ca(s) + O2(g) --> 2 CaO(s) H = -1270.2 kJ
C(s) + O2(g) --> CO2(g) H = -393.5 kJ
2 Ca(s) + 2 C(s) + 3 O2(g) -->2 CaCO3(s) H = -2413.8 kJ

a. -4077.3 kJ
b. -750.1 kJ
c. -178.3 kJ
d. +350.2 kJ
e. +2870.6 kJ



How do you solve??

Answer 1

This one is pretty simple....
1st eq. = flip (so make value positive) and divide by 2
2nd eq. = flip (make value positive)
3rd eq. = leave as is but divide by 2

so the answer is
1270.2kj/2 +393.5 kJ + -2413.8kJ/2
= 635.1kJ+393.5 kJ-1206.9kJ
=-178.3
= (c)

Answer 2

One looks at what is being created and what is being distroyed. Thermochemical equations are given in terms of things being created. Therefore, the sign has to be reversed for anything distroyed. Hence:

CO2 (destroyed) = +393.5 kJ
2CaO (destroyed) = 1270 kJ, so one of them = 635 kJ
2CaCO3 (created) = -2413.8 kJ, so one = -1206.5 kJ
---------------------------------------------------------------------------
total = 393 + 635 - 1206 = -178 kJ

The answer is negative, which means the reaction is spontaneous. This seems confusing, but thermodynamic calculations involve the change in energy state calculated by subtracting "before" from "after". Spontaneous reactions involve substances with more energy loosing it as products are formed. This spontaneous decrease in energy is indicated by a negative answer.

Answer 3

Half the first equation, minus the second, + half the third.

Now you can work it out!