As an example, let’s say we want to find the enthalpy of reaction for the formation of water from hydrogen and oxygen: 2H2 (Hydrogen) + O2 (Oxygen) → 2H2O (Water). In this equation, H2 and O2 are the reactants and H2O is the product.
In our water example, our reactants are hydrogen and oxygen gases, which have molar masses of 2g and 32 g, respectively. Since we used 2 moles of hydrogen (signified by the “2” coefficient in the equation next to H2) and 1 mole of oxygen (signified by no coefficient next to O2), we can calculate the total mass of the reactants as follows:2 × (2g) + 1 × (32g) = 4g + 32g = 36g
Note that if your equation has multiple products, you’ll need to perform the enthalpy calculation for the component reaction used to produce each product, then add them together to find the enthalpy for the entire reaction. In our example, the final product is water, which has a specific heat of about 4. 2 joule/gram °C.
For our example, let’s say that our reaction was 185K at its very start but had cooled to 95K by the time it finished. In this case, ∆T would be calculated as follows:∆T = T2 – T1 = 95K – 185K = -90K
For our example problem, we would find the enthalpy of reaction as follows:∆H = (36g) × (4. 2 JK-1 g-1) × (-90K ) = -13,608 J
In our example, our final answer is -13608 J. Since the sign is negative, we know that our reaction is exothermic. This makes sense — H2 and O2 are gasses, while H2O, the product, is a liquid. The hot gasses (in the form of steam) have to release energy into the environment in the form of heat to cool to the point that they can form liquid water, meaning that the formation of H2O is exothermic.
For example, let’s consider the reaction H2 + F2 → 2HF. In this case, the energy required to break the H atoms in the H2 molecule apart is 436 kJ/mol, while the energy required for F2 is 158 kJ/mol. Finally, the energy needed to form HF from H and F is = -568 kJ/mol. We multiply this by 2 because the product in the equation is 2HF, giving us 2 × -568 = -1136 kJ/mol. Adding these all up, we get: 436 + 158 + -1136 = -542 kJ/mol.
For example, let’s consider the reaction C2H5OH + 3O2 → 2CO2 + 3H2O. In this case, we know the enthalpies of formation for the following reactions:C2H5OH → 2C + 3H2 + 0. 5O2 = 228 kJ/mol2C + 2O2 → 2CO2 = -394 × 2 = -788 kJ/mol 3H2 + 1. 5 O2 → 3H2O = -286 × 3 = -858 kJ/mol Since we can add these equations up to get C2H5OH + 3O2 → 2CO2 + 3H2O, the reaction we’re trying to find the enthalpy for, we can simply add up the enthalpies of the formation reactions above to find the enthalpy of this reaction as follows: 228 + -788 + -858 = -1418 kJ/mol.
In the example above, notice that the formation reaction we use for C2H5OH is backwards. C2H5OH → 2C + 3H2 + 0. 5O2 shows C2H5OH breaking down, not being formed. Because we turned the equation around in order to get all of the products and reactants to cancel properly, we reversed the sign on the enthalpy of formation to give us 228 kJ/mol. In reality, the enthalpy of formation for C2H5OH is -228 kJ/mol.
For this experiment, you’ll want a fairly small container. We’ll be testing the enthalpy-altering effects of Alka-Seltzer on water, so the less water used, the more obvious the temperature change will be.
Let’s say that we measure the temperature of the water and find that it’s exactly 10 degrees C. In a few steps, we’ll use this sample temperature reading to demonstrate the principals of enthalpy.
For our example experiment, let’s say that the temperature of the water is 8 degrees C after the tablet has finished fizzing.
In our example experiment, the temperature of the water fell two degrees after adding the Alka-Seltzer. This is consistent with the sort of mildly endothermic reaction we’d expect.
