Suppose 23.6 g of ClF3(g) and 40 g of Li(s) are mixed and allowed to react at atmospheric pressure and 25 C until one of the reactants is used up, producing LiCl(s) and LiF(s). Using the NIST value for the enthalpy of formation of ClF3 (webbook.nist.gov), calculate the enthalpy change for this reaction, keeping in mind that this is a negative number for an exothermic process.
To calculate the enthalpy change for the reaction mentioned, we need to use the thermochemical equation for the reaction and apply the law of conservation of mass.
The given reaction is:
ClF3(g) + 3Li(s) -> LiCl(s) + LiF(s)
To calculate the enthalpy change (ΔH) for the reaction, we will first determine the limiting reactant, which is the reactant that will be completely used up in the reaction. The reactant that produces the least moles of product is the limiting reactant.
We can start by calculating the number of moles for each reactant using their molar masses:
Molar mass of ClF3 = 83.45 g/mol
Molar mass of Li = 6.94 g/mol
Number of moles of ClF3 = mass / molar mass = 23.6 g / 83.45 g/mol
Number of moles of ClF3 = 0.283 mol
Number of moles of Li = mass / molar mass = 40 g / 6.94 g/mol
Number of moles of Li = 5.76 mol
According to the balanced equation, 1 mole of ClF3 reacts with 3 moles of Li. Therefore, the ratio of moles is 1:3.
Since we have 0.283 moles of ClF3 and 5.76 moles of Li, we can see that there is an excess of Li. To find the amount of Li that reacts, we need to use the stoichiometry of the balanced equation.
The ratio of Li to ClF3 is 3:1. Therefore, to react with 0.283 moles of ClF3, we need (0.283 mol ClF3) x (3 mol Li / 1 mol ClF3) = 0.849 moles of Li. Since we have an excess of Li, only 0.283 moles of Li will react.
Now, we can calculate the enthalpy change using the enthalpy of formation values.
The enthalpy change for a reaction can be calculated using the equation:
ΔHrxn = ΣnΔHf(products) - ΣnΔHf(reactants)
Where ΣnΔHf represents the sum of the products or reactants' enthalpies of formation, each multiplied by their respective number of moles.
The values of the enthalpies of formation for LiCl and LiF can be found in the NIST database: webbook.nist.gov.
Given that ΔHf° (enthalpy of formation) of LiCl(s) = -408.4 kJ/mol and ΔHf° of LiF(s) = -618.0 kJ/mol, we can now substitute these values into the enthalpy change equation.
ΔHrxn = (1 mol LiCl) x (-408.4 kJ/mol) + (1 mol LiF) x (-618.0 kJ/mol)
ΔHrxn = -408.4 kJ/mol - 618.0 kJ/mol
ΔHrxn = -1026.4 kJ/mol
Therefore, the enthalpy change for this reaction is -1026.4 kJ/mol. Since it is negative, it indicates that the reaction is exothermic.