Using Hess\'s Law, show how the three reactions in the BACKGROUND section of the

Question

Using Hess's Law, show how the three reactions in the BACKGROUND section of the lab can be combined to obtain the standard molar enthalpy of formation of MgO(s), Mg(s) + 1/2 O2(g)=MgO(s). You won't have actual numbers to use here, you are just showing how the equations for the reactions and the enthalpy changes for those reactions can be manipulated to give the desired reaction. Three reactions from background: 1. Mg(s)+2HCl(aq)=MgCl2(aq)+H2(g) 2. MgO(s)+2HCl(aq)=MgCl2(aq)+H2O(l) 3. H2(g)+1/2O2(g)=H2O(l) with H° = -57.7 kJ

Explanation / Answer

Get H for

Mg(s) + 1/2 O2(g)=MgO(s)

applying hess law with the next equations:;

1. Mg(s)+2HCl(aq)=MgCl2(aq)+H2(g) H1

2. MgO(s)+2HCl(aq)=MgCl2(aq)+H2O(l) H2

3. H2(g)+1/2O2(g)=H2O(l) with H° = -57.7 kJ = H3

Invert (2)

1. Mg(s)+2HCl(aq)=MgCl2(aq)+H2(g) H1

2. MgCl2(aq)+H2O(l) = MgO(s)+2HCl(aq) -H2

3. H2(g)+1/2O2(g)=H2O(l) with H° = -57.7 kJ = H3

Add equations (1)+(2)+(3)

Mg(s)+2HCl(aq) + MgCl2(aq)+H2O(l) + H2(g)+1/2O2(g) = MgCl2(aq)+H2(g) + MgO(s)+2HCl(aq) +H2O(l)

H = H1-H2+H3

Cancel common terms

Mg(s)+2HCl(aq) + MgCl2(aq)+H2O(l) + H2(g)+1/2O2(g) = MgCl2(aq)+H2(g) + MgO(s)+2HCl(aq) +H2O(l)

Mg(s)+1/2O2(g) =  MgO(s)

which is what we wanted... overal Hrxn = H1-H2 + H3

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