Question number 1.
Hess’s Law states that the change of enthalpy is the same whether the reaction occurs in one step
or many steps. In other words all of the enthalpies of the partial reactions need to add up to be
the enthalpy of the total reaction. So,
Given the complete equation:
Mg(s) + ½ O2(g) → MgO(s) with enthalpy ΔHrxn
And given the three partial equations:
Eqn1) Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
with enthalpy ΔH1
Eqn2) MgO(s) + 2 HCl(aq) → MgCl2(aq) + H2O(l) with enthalpy ΔH2
Eqn3) H2(g) + ½ O2(g) → H2O(l) with enthalpy ΔH3= -285.8 kJ
We can see that if we reverse Eqn 2 to read :
MgCl2(aq) + H2O(l) → MgO(s) + 2 HCl(aq)
And change the enthalpy to - ΔH2. We are left with our complete equation because the HCl, H2,
and MgCl all cancel out. And since Hess Law is just the addition of the enthalpies the new
equation reads:
ΔHrxn = ΔH1 + ΔH3 - ΔH2
Question number 2.
Given the explanation in the lab for each variable for the equation -n ΔHo = Ccal Δ T +
(m)(c)(ΔT)., we can assign each with a value:
n= moles of the limiting reactant. Since NaOH and HCl are stoichiometrically equivalent, the
one with the least amount of moles is going to be the limiting reactant. To find moles of each we
multiply molarity by volume in liters
for HCL: (.909M)*(.0253L) = .023 moles HCL
for NaOH: (1.09M)*(.0241L) = .026 moles NaOH therefore HCL is the limiting reactant and
n=.023 moles
ΔHo = -57.7 kJ/mole (this was given) when switched to J 57700 Joules
Δ T = the change in temperature so (Tfinal- Tinitial) which is (24.51ºC-22.92ºC) = 1.59ºC
m = the mass of the solution, which we have to solve for using the density of 1.00g/mL =
mass/volume. The volume= 25.3mL+24.1mL. So therefore the mass = 49.4 g
c= specific heat of the solution. This was given in the lab book as 4.025 J/(g°C).
So given the equation : -n ΔHo = Ccal Δ T + (m)(c)(ΔT) we can plug in the values above:
-(.023 moles) (-57700J/mol) = Ccal (1.59ºC) + (49.4g)(4.025J/(gºC))(1.59ºC)
which when simplified is :
1327.1 J = Ccal (1.59ºC) + 316.15J
1,010.95 J = Ccal (1.59ºC)
Ccal =1,010.95J/1.59 ºC= 635.82 J/ºC
Question number 3.
The reaction is :
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
To solve for the limiting reactant, we need to see which will stoichiometrically produce less
product. Using MgCl2 as our product:
For Mg:
.475gMg(1 mole/24.305g)= .0159 moles Mg(1 mole MgCl2/1 mole Mg) = .0159 moles MgCl2
For HCl:
.941M*.058L= .055 moles HCl (1 mole MgCl2/ 2 moles HCl) = .027 moles MgCl2
Therefore Mg(s) is the limiting reactant.
Now using the equation we used in question 2: -n ΔHo = Ccal Δ T + (m)(c)(ΔT)
n = .0159 moles Mg(s) determined in the last part.
Ccal= 13.0 J/ºC
Δ T = 14.9ºC
m = 100 g (using the density of 1g/mL and information from the lab see pg. 7)
c = 3.862 J/(g°C) given in the lab
So now plugging those values into the equation: -n ΔHo = Ccal Δ T + (m)(c)(ΔT)
-(.0159moles Mg) ΔHo = (13.0 J/ºC)(14.9ºC) + (100 g)(3.862 J/(g°C))(14.9ºC)
-0.159moles* ΔHo = 193.7J + 5754.38J
-0.159moles* ΔHo = 5948.08J
ΔHo = 37.41 kJ/mol
Question number 4.
If you remove 2.97 kJ of heat from 235. g of water at 49.5oC, what temperature change would
you observe in the water and what would be the final temperature of the water? The specific heat
capacity of water is 4.184 J/(g oC).
For this problem, since we know the exact composition, we can use the formula q=mc Δ T
where:
q = heat removed = 2.97kJ
m = mass which would be 235g
c = specific heat of 4.184J/(g ºC)
so using our equation
2,970J= (235g)(4. 184J/(g ºC))( Δ T)
Δ T = 2,970J/ (235g)(4. 184J/(g ºC)) = 3.02ºC
So remember
Δ T = (Tfinal – Tinitial ) so if Δ T = 3.02 and Tinitial = 49.5ºC
3.02ºC = Tfinal – 49.5ºC
Tfinal = 52.52 ºC