Natural diamonds are mined from sites around the world. However, the price of natural diamonds is carefully controlled, so other sources of diamonds are being explored. Several different methods for producing synthetic diamonds are available, usually involving treating carbon at very high temperatures and pressures. The diamonds produced are now of high quality, but are primarily used in industrial applications. Diamonds are one of the hardest materials available and are widely used for cutting and grinding tools. Calculating Heat of Reaction from Heat of Formation An application of Hess's law allows us to use standard heats of formation to indirectly calculate the heat of reaction for any reaction that occurs at standard conditions. An enthalpy change that occurs specifically under standard conditions is called the standard enthalpy (or heat) of reaction and is given the symbol \(\Delta H^\text{o}\). The standard heat of reaction can be calculated by using the following equation. \[\Delta H^\text{o} = \sum n \Delta H^\text{o}_\text{f} \: \text{(products)} - \sum n \Delta H^\text{o}_\text{f} \: \text{(reactants)}\nonumber \] The symbol \(\Sigma\) is the Greek letter sigma and means "the sum of". The standard heat of reaction is equal to the sum of all the standard heats of formation of the products minus the sum of all the standard heats of formation of the reactants. The symbol "\(n\)" signifies that each heat of formation must first be multiplied by its coefficient in the balanced equation. Table \(\PageIndex{1}\): Standard Heats of Formation of Selected Substances Table \(\PageIndex{1}\): Standard Heats of Formation of Selected Substances.1 Table \(\PageIndex{1}\): Standard Heats of Formation of Selected Substances.2 Table \(\PageIndex{1}\): Standard Heats of Formation of Selected Substances.3 Substance \(\Delta H^\text{o}_\text{f}\) \(\left( \text{kJ/mol} \right)\) Substance \(\Delta H^\text{o}_\text{f}\) \(\left( \text{kJ/mol} \right)\) \(\ce{Al_2O_3} \left( s \right)\) -1669.8 \(\ce{H_2O_2} \left( l \right)\) -187.6 \(\ce{BaCl_2} \left( s \right)\) -860.1 \(\ce{KCl} \left( s \right)\) -435.87 \(\ce{Br_2} \left( g \right)\) 30.91 \(\ce{NH_3} \left( g \right)\) -46.3 \(\ce{C} \left( s, graphite \right)\) 0 \(\ce{NO} \left( g \right)\) 90.4 \(\ce{C} \left( s, diamond \right)\) 1.90 \(\ce{NO_2} \left( g \right)\) 33.85 \(\ce{CH_4} \left( g \right)\) -74.85 \(\ce{NaCl} \left( s \right)\) -411.0 \(\ce{C_2H_5OH} \left( l \right)\) -276.98 \(\ce{O_3} \left( g \right)\) 142.2 \(\ce{CO} \left( g \right)\) -110.5 \(\ce{P} \left( s, white \right)\) 0 \(\ce{CO_2} \left( g \right)\) -393.5 \(\ce{P} \left( s, red \right)\) -18.4 \(\ce{CaO} \left( s \right)\) -635.6 \(\ce{PbO} \left( s \right)\) -217.86 \(\ce{CaCO_3} \left( s \right)\) -1206.9 \(\ce{S} \left( rhombic \right)\) 0 \(\ce{HCl} \left( g \right)\) -92.3 \(\ce{S} \left( monoclinic \right)\) 0.30 \(\ce{CuO} \left( s \right)\) -155.2 \(\ce{SO_2} \left( g \right)\) -296.1 \(\ce{CuSO_4} \left( s \right)\) -769.86 \(\ce{SO_3} \left( g \right)\) -395.2 \(\ce{Fe_2O_3} \left( s \right)\) -822.2 \(\ce{H_2S} \left( s \right)\) -20.15 \(\ce{H_2O} \left( g \right)\) -241.8 \(\ce{SiO_2} \left( s \right)\) -859.3 \(\ce{H_2O} \left( l \right)\) -285.8 \(\ce{ZnCl_2} \left( s \right)\) -415.89 Example \(\PageIndex{1}\) Calculate the standard heat of reaction \(\left( \Delta H^\text{o} \right)\) for the reaction of nitrogen monoxide gas with oxygen to form nitrogen dioxide gas. Solution Step 1: List the known quantities and plan the problem. Known \(\Delta H^\text{o}_\text{f}\) for \(\ce{NO} \left( g \right) = 90.4 \: \text{kJ/mol}\) \(\Delta H^\text{o}_\text{f}\) for \(\ce{O_2} \left( g \right) = 0\) (element) \(\Delta H^\text{o}_\text{f}\) for \(\ce{NO_2} \left( g \right) = 33.85 \: \text{kJ/mol}\) Unknown First write the balanced equation for the reaction. Then apply the equation to calculate the standard heat of reaction from the standard heats of formation. Step 2: Solve. The balanced equation is: \(2 \ce{NO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)\) Applying the equation from the text: \[\begin{align*} \Delta H^\text{o} &= \left[ 2 \: \text{mol} \: \ce{NO_2} \left( 33.85 \: \text{kJ/mol} \right) \right] - \left[ 2 \: \text{mol} \: \ce{NO} \left( 90.4 \: \text{kJ/mol} \right) + 1 \: \text{mol} \: \ce{O_2} \left( 0 \: \text{kJ/mol} \right) \right] \\ &= -113 \: \text{kJ} \end{align*}\nonumber \] The standard heat of reaction is \(-113 \: \text{kJ}\nonumber \] Step 3: Think about your result. The reaction is exothermic, which makes sense because it is a combustion reaction and combustion reactions always release heat. Summary An enthalpy change that occurs specifically under standard conditions is called the standard enthalpy (or heat) of reaction and is given the symbol \(\Delta H^\text{o}\). Standard heats of reaction can be calculated from standard heats of formation.