Calculating the Heat of Reaction from the Heats of Formation

Let's find the heat of the reaction:

CaCO_{3 (s)} + 2 NaOH _(s) → Ca(OH)_{2 (s)} + Na₂CO_{3 (s)}

The formula we use fir a calculation like this is:  ΔH_rxn = ΣΔH_{f products} – ΣΔH_{f reactants}

So, the heat of reaction should be calculated like this:

ΔH_rxn = [ ΔH_{f Ca(OH)2} + ΔH_{f Na2CO3}] - [ ΔH_{f CaCO3} + 2(ΔH_{f NaOH})]

Notice that the 2 from the balanced reaction shows up in the math! Substituting in the values from the table gives:

ΔH_rxn= [-986.09 kJ/mol + -1130.77 kJ/mol] - [-1207.6 kJ/mol + 2(-425.93 kJ/mol)]

solving this gives us:

ΔH_rxn= [-2116.86 kJ/mol] - [-2059.46 kJ/mol] = -57.4 kJ/mol