# Exo/Endo Energy Cycle

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#1
How do I answer question 13b) i)
I know there are two Born-Haber Cycles cycles for exothermic and endothermic reactions
How do I know if it is exothermic or endothermic if I actually have to work out the enthalpy of solution?

Question: http://pmt.physicsandmathstutor.com/...%20Entropy.pdf
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3 years ago
#2
(Original post by esmeralda123)
How do I answer question 13b) i)
I know there are two Born-Haber Cycles cycles for exothermic and endothermic reactions
How do I know if it is exothermic or endothermic if I actually have to work out the enthalpy of solution?

Question: http://pmt.physicsandmathstutor.com/...%20Entropy.pdf
if an enthalpy change is + then the reaction is endothermic
otherwise its exothermic if enthalpy change is -
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#3
(Original post by dip0)
if an enthalpy change is + then the reaction is endothermic
otherwise its exothermic if enthalpy change is -
Yes, but where is the enthalpy change stated in the question?
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3 years ago
#4
(Original post by esmeralda123)
Yes, but where is the enthalpy change stated in the question?
Use the data given below to calculate the standard enthalpy of solution of CaCl2.
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#5
(Original post by dip0)
Use the data given below to calculate the standard enthalpy of solution of CaCl2.
Yes this is what I don't get.
I can only decide if a Born Haber Cycle is endothermic or exothermic with the sign of the standard netherlpy of solution
But if I have to calculate it, how I am supposed to draw the cycles to work it out?
Or are you suggesting there is a general formula?
If there is, what is the point of the endo/exo Born Haber Cycles?
0
3 years ago
#6
(Original post by esmeralda123)
Yes this is what I don't get.
I can only decide if a Born Haber Cycle is endothermic or exothermic with the sign of the standard netherlpy of solution
But if I have to calculate it, how I am supposed to draw the cycles to work it out?
Or are you suggesting there is a general formula?
If there is, what is the point of the endo/exo Born Haber Cycles?
there is a general formula of working it out using a thermochemical cycle.

we are interested in this:
CaCl2(s) + (aq) => Ca2+(aq) + 2Cl-(aq)

1) convert the solid calicum chloride into its gaseous ions. ( = lattice dissociation enthalpy)

2a) hydrate the calcium ion
2b) hydrate chloride ions (= enthalpy of hydration of chloride X 2)

to work out enthalpy change just do:
1 + 2a + 2b
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#7
(Original post by dip0)
there is a general formula of working it out using a thermochemical cycle.

we are interested in this:
CaCl2(s) + (aq) => Ca2+(aq) + 2Cl-(aq)

1) convert the solid calicum chloride into its gaseous ions. ( = lattice dissociation enthalpy)

2a) hydrate the calcium ion
2b) hydrate chloride ions (= enthalpy of hydration of chloride X 2)

to work out enthalpy change just do:
1 + 2a + 2b
Thank you.
Does this apply for exothermic and endothermic reactions ?
Also, the last part of the question about solubility, why does it decrease. I don't even understand the whole answer to the question.
0
3 years ago
#8
its a reversible reaction.

CaCl2 + (aq) <=> Ca2+ + 2Cl-
forward reaction is exothermic (we know that because the enthalpy change is - )
hence backwards reaction is endothermic
its just an application of le chateliers principle.
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3 years ago
#9
(Original post by esmeralda123)
How do I answer question 13b) i)
I know there are two Born-Haber Cycles cycles for exothermic and endothermic reactions
How do I know if it is exothermic or endothermic if I actually have to work out the enthalpy of solution?

(i) Use the data given below to calculate the standard enthalpy of solution of CaCl2.
Lattice dissociation enthalpy of CaCl2 = +2255 kJ mol–1Hydration enthalpy of calcium ions = –1650 kJ mol–1
Hydration enthalpy of chloride ions = –384 kJ mol–1
Step 1 is to write out the relevant equations that represent the processes above.

1. CaCl2(s) --> Ca2+(g) + 2Cl- (g)....... +2255 kJ
2. Ca2+(g) --> Ca2+(aq) ...................................... -1650 kJ
3. Cl-(g) --> Cl-(aq)............................................... -384 kJ

multiply through by 2 in equation 3 because you need two chloride ions:

3. 2Cl-(g) --> 2Cl-(aq)............................................... -768 kJ

1. CaCl2(s) --> Ca2+(g) + 2Cl- (g)....... +2255 kJ
2. Ca2+(g) --> Ca2+(aq) ...................................... -1650 kJ
4. CaCl2(s) -->Ca2+(aq) + 2Cl- (g)...... +605 kJ

now add equation 3 & 4
4. CaCl2(s) -->Ca2+(aq) + 2Cl- (g)......... +605 kJ
3. 2Cl-(g) --> 2Cl-(aq)............................................... -768 kJ
CaCl2(s) --> Ca2+(aq) + 2Cl-(aq)............ -163 kJ
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