Heat released in the reaction can be calculated by the formula:
Q = mc∆T
Whereby;
m = Mass of the solution
c = Specific heat of water (4.2 Jmol-1°C-1)
∆T = Change in temperature
Enthalpy change of the reaction can be determined by the formula:-
∆H = Amount of heat released/absorbed, Q
Number of moles of reactant, n
Calculation:
Heat change, Q = mc∆T
= 50g × 4.2 Jmol-1°C-1 × 13.5 °C
= 2835 J
∆Hrxn = -2835_
0.0118
= -240.25 kJmol-1
Uncertainties:
Mass = = 1.0 %
Temperature = = 7.4 %
No of mole = = 0.85 %
Total = 9.25%
∆Hrxn = -240.25 kJmol-1 ± 9.25 %
Part B:-
Mass of magnesium oxide = 0.4985 ± 0.0001g
Volume of 0.5 M of HCl = 50.0 ± 0.5 cm3
Table 3 shows the initial and highest temperature for reaction between MgO and HCl
To calculate the change in temperature, the following equation will be used:-
Table 4 shows the change in temperature for reaction between MgO and HCl
Equation for part B:-
MgO (s) + 2HCl (aq) → MgCl2 (aq) + H2O (l)
Mole of magnesium oxide Mole of Hydrochloric acid
= mass = MV
RMM 1000
= 0.4985g = 50x0.5
(24.3+16) 1000
= 0.0124 mol = 0.025 mol
Thus, the limiting reagent is magnesium oxide
Calculation:
Heat change, Q = mc∆T
= 50g × 4.2 Jmol-1°C-1 × 2 °C
= 420 J
∆Hrxn = -420_
0.0124
= -33.87 kJmol-1
Uncertainties:
Mass = = 1.0 %
Temperature = = 50.0 %
No of mole = = 8.1%
Total = 59.1%
∆Hrxn = -33.87 kJmol-1 ± 59.1 %
Since the reactions are exothermic, the ∆Hrxn for each reaction is negative. Thus
Hess’s Law states that enthalpy change is independent of the route taken. The enthalpy change for the reaction will be equal to the sum of the enthalpy change of each step. Thus for both reactions below
The reaction which ∆Hrxn is being investigated:
Mg(s) + H2O(l) → MgO(s) + H2(g)
Information obtained from the experiment:
-
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) ∆Hrxn = -240.25 kJmol-1
-
MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) ∆Hrxn = -33.87 kJmol-1
By reversing equation 2:
3. MgCl2(aq) + H2O(l) → MgO(s) + 2HCl(aq) ∆Hrxn = +33.87 kJmol-1
To obtain the desired equation, equation 1 + equation 3:
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) ∆Hrxn = -240.25 kJmol-1 ± 9.25%
MgCl2(aq) + H2O(l) → MgO(s) + 2HCl(aq) ∆Hrxn = +33.87 kJmol-1 ± 59.1%
Mg(s) + H2O(l) → MgO(s) + H2(g) ∆Hrxn = -206.38 kJmol-1 ± 68.35%
∆Hrxn = -206.38 kJmol-1 ± 16.12 %
For the theoretical value:
1. Mg(s) + (g) MgO(s) ∆Hrxn = -601.8 kJmol-1
2. H2(g) + (g) H2O(l) ∆Hrxn = -285.8 kJmol-1
By reversing equation 2:
3. H2O(l) H2(g) + (g) ∆Hrxn = +285.8 kJmol-1
The theoretical value of the reaction, equation 1 + equation 3:
Mg(s) + (g) MgO(s) ∆Hrxn = -601.8 kJmol-1
H2O(l) H2(g) + (g) ∆Hrxn = +285.8 kJmol-1
Mg(s) + H2O(l) MgO(s) + H2(g) ∆Hrxn = -316.0 kJmol -1
∆Hrxn obtained from this experiment = -206.38 kJmol-1
The theoretical value of ∆Hrxn = -316.0 kJmol -1
Percentage error = Theoretical value – Experimental value
Theoretical value
=
= 34.69%
Reference:
Chemistry: A Central Science by Brown, Le May and Bursten
