Question

Using the equations Ca (s)2 O (g) Сао (s) 2 H2 (g) Determine the enthalpy for the reaction Са (s) + 2 H0 () — Са(ОН)2 (s) + H2 (9). CaO (s) AH° = -635 kJ/mol -64 kJ/mol H2O ()Ca(OH)2 (s) AH O2 (g)2 H2O (I) AH° = -572 kJ/mol kJ/mol 1 2 3 4 6 C 5 7 8 9 +/- 0 x 100 LO 11.01 g of MgSO4 is placed into 100.0 mL of water. The water's temperature increases by 6.7°C. Calculate AH, in kJ/mol, for the dissolution of MgSO4. (The specific heat of water is 4.18 J/g"C and the density of the water is 1.00 g/mL). You can assume that the specific heat of the solution is the same as that of water. kJ/mol 1 2 4 5 C 7 9 +- 0 x 100 LO

Answer 1

1) First, you have three reaction

AH 635 Kj/mol Ca(s)1/202Cao(s)

CaO (s)H2O(1)Ca(OH)2(s) ДН — — 64 Кj/mol

2H29)O2(g) 2H20() AH-572 Kj/mol

with the product

Ca(s)2H2O(I)Ca(OH)2(s) H2(g)

You need to follow HESS LAW, where you fix your principal reactions to get the same order just like the product.

So cheching each reaction, the first reaction you se Ca(s) is a reactant and in your final reaction keeps as a reactant, then the first reaction keeps the same

(1)

AH 635 Kj/mol Ca(s)1/202Cao(s)

In the second reaction, check H2O, which is a reactant meanwhile in you final reaction is a reactant, so there is no change with this reaction and the same with Ca(OH)2 which is a product

(2)

CaO (s)H2O(1)Ca(OH)2(s) ДН — — 64 Кj/mol

Finally in the third reaction, H2 is a reactant, but in your final reaction is a product, so you change the order of the third reaction, and as you change the order, it is important to change the value of H, passing from -572 to +572. Finally, you'll see that in your final reaction H2 does not have 2moles, that is why in your third reaction you need to divide by 2 and get the same number of moles

(2H29) + О2(9)н 2H20 () /2 АН -572/2 Кjтої _

2H2O()2H2(g) +O2(g))/2 AH= +572/2 Kj/mol _

(3)

H2O()H2(g) 1/202(g) AH +286 Kj/mol

Now, when you put together this 3 reactions, there are substances that will be eliminated between reactants and products, such as O2, CaO, getting your principal reaction.

And for AH, just do the mathematical operation where H=-635-64+286=-413 kj/mol

Ca(s)2H20(1)Ca(OH)2(s) H2(g) AH=413kj/mol

2) First you need to calculate the heat change of water, with

, where m=mass of water, T=change in temperature, c=specific heat

g =mATc

for mass of water = Volume of water * density of water

HO 100 ml1g/ml 100 grams

Now, calculate heat change.

g100grams 6.7°C* 4.18 J/g* C 2800.6J

Then, the enthalpy change is for the dissolution of MgSO4(mass) into water, and as enthalpy change is given in Joules/moles, you need to know the number of moles of MgSO4 used in this dissolution

11.01 grams 0.09147 moles 120.36 grams/moles

Finally, make the relation between heat transfer of water into the number of moles of MgSO4

$\Delta H=\frac{2800.6\, J}{0.09147\, mol}=30617.68\, J/mol$

Since this enthalpy is issued from the reaction, represented as energy that is losing, so the value is changed

ТK) -30617.68 J/mоol * 1000J —30.617 Кj/mой