alculate the enthalpy of the following reaction: (3 marks)2Mg + O2 → 2MgOGiven:Mg + 2HCl → MgCl2 + H2 ΔH = -21.0 kJMgO + 2HCl → MgCl2 + H2O ΔH = -33.5 kJH2 + ½ O2 → H2O ΔH = -76.0 kJ
Question
alculate the enthalpy of the following reaction: (3 marks)2Mg + O2 → 2MgOGiven:Mg + 2HCl → MgCl2 + H2 ΔH = -21.0 kJMgO + 2HCl → MgCl2 + H2O ΔH = -33.5 kJH2 + ½ O2 → H2O ΔH = -76.0 kJ
Solution
To calculate the enthalpy of the reaction 2Mg + O2 → 2MgO, we can use Hess's Law, which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which a reaction can be divided.
The given reactions are:
- Mg + 2HCl → MgCl2 + H2 ΔH = -21.0 kJ
- MgO + 2HCl → MgCl2 + H2O ΔH = -33.5 kJ
- H2 + ½ O2 → H2O ΔH = -76.0 kJ
We need to manipulate these reactions to get the desired reaction (2Mg + O2 → 2MgO).
First, multiply reaction 1 by 2 to get 2Mg + 4HCl → 2MgCl2 + 2H2. The enthalpy change is -21.0 kJ x 2 = -42.0 kJ.
Second, multiply reaction 2 by 2 to get 2MgO + 4HCl → 2MgCl2 + 2H2O. The enthalpy change is -33.5 kJ x 2 = -67.0 kJ.
Third, multiply reaction 3 by 2 to get 2H2 + O2 → 2H2O. The enthalpy change is -76.0 kJ x 2 = -152.0 kJ.
Now, add the first and third reactions and subtract the second reaction:
(2Mg + 4HCl + 2H2 + O2) - (2MgO + 4HCl) = (2MgCl2 + 2H2 + 2H2O) - (2MgCl2 + 2H2O)
This simplifies to the desired reaction: 2Mg + O2 → 2MgO.
The enthalpy change of this reaction is the sum of the enthalpy changes of the individual steps: -42.0 kJ + (-152.0 kJ) - (-67.0 kJ) = -127.0 kJ.
So, the enthalpy of the reaction 2Mg + O2 → 2MgO is -127.0 kJ.
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