# Bohr-haber cycles - a level chemistry

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#1
I got a decimal for (ii) so not sure if I have gone wrong anywhere. Also, for (iii) it asked to calculate the lattice enthalpy but I calculated it in (ii) which confused me.
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2 months ago
#2
(Original post by dnejfn)
I got a decimal for (ii) so not sure if I have gone wrong anywhere. Also, for (iii) it asked to calculate the lattice enthalpy but I calculated it in (ii) which confused me.

(i) The lower dotted line is the elements in their standard states, i.e. Mg(s) + Br2(l)

The upper dotted line is Mg2+(g) + 2Br(g)

(ii)
You need the enthalpy of hydration of bromide ions, i.e. Br-(g) ==> Br-(aq)
There is no lattice enthalpy step in the cycle.

You do need to add up (magnitude) all of the LHS and it should equal all of the RHS (just one way to do it.
The other way is to go around the cycle in the other direction to the step that you are trying to determine (bottom right)

186 + 525 + 2(112) + 148 + 736 + 1450 = (2 x dH(hydration) Br-) + 1926 + (2 x 325)

Rearrange

(2 x dH(hydration) Br-) = 693

Hydration enthalpy of bromide ions = -346.5 kJ (hydration is always exothermic)
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#3
(Original post by charco)

(i) The lower dotted line is the elements in their standard states, i.e. Mg(s) + Br2(l)

The upper dotted line is Mg2+(g) + 2Br(g)

(ii)
You need the enthalpy of hydration of bromide ions, i.e. Br-(g) ==> Br-(aq)
There is no lattice enthalpy step in the cycle.

You do need to add up (magnitude) all of the LHS and it should equal all of the RHS (just one way to do it.
The other way is to go around the cycle in the other direction to the step that you are trying to determine (bottom right)

186 + 525 + 2(112) + 148 + 736 + 1450 = (2 x dH(hydration) Br-) + 1926 + (2 x 325)

Rearrange

(2 x dH(hydration) Br-) = 693

Hydration enthalpy of bromide ions = -346.5 kJ (hydration is always exothermic)
is (iii) right? I got 1383
0
2 months ago
#4
(Original post by dnejfn)
is (iii) right? I got 1383

MgBr2(s) ==> Mg2+(g) + 2Br-(g)

As you are working with solutions in parts (i) and (ii) you should get from MgBr2(s) to Mg2+(g) + 2Br-(g) via MgBr2(aq)

MgBr2(s) ==> MgBr2(aq) ............... ΔH = -186

Then add in the reverse of the hydration enthalpies of the two ions

Mg2+(aq) ==> Mg2+(g) ...........ΔH = +1926 kJ

2Br-(aq) ==> 2Br-(g) ........... ΔH = +693kJ

Add all three together = +2433 kJ
Last edited by charco; 2 months ago
0
3 weeks ago
#5
(Original post by charco)

MgBr2(s) ==> Mg2+(g) + 2Br-(g)

As you are working with solutions in parts (i) and (ii) you should get from MgBr2(s) to Mg2+(g) + 2Br-(g) via MgBr2(aq)

MgBr2(s) ==> MgBr2(aq) ............... ΔH = -186

Then add in the reverse of the hydration enthalpies of the two ions

Mg2+(aq) ==> Mg2+(g) ...........ΔH = +1926 kJ

2Br-(aq) ==> 2Br-(g) ........... ΔH = +693kJ

Add all three together = +2433 kJ
Isn't lattice enthalpy always negative?
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3 weeks ago
#6
(Original post by AltAccount82232)
Isn't lattice enthalpy always negative?
There are two different definitions. Different examination boards use one or the other:

The lattice dissociation enthalpy is always positive. NaCl(s) ==> Na+(g) + Cl-(g)
The lattice enthalpy is always negative. Na+(g) + Cl-(g) ==> NaCl(s)
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