It's a molar heat reaction lab.
0.07g Mg
Mg+2HCl->MgCl2+H2
What is the actual value for the heat of reaction based on the enthalpy's of formation?
HfHCl= -118.53 kJ/mole
HfMgCl2= -774 kJ/mole
dHrxn = (n*dHf MgCl2) - (dHf HCl) = ?
To calculate the actual value for the heat of reaction based on the enthalpies of formation, you need to use the heat of formation values for the reactants and products involved in the reaction.
The enthalpy change for the reaction can be calculated using the equation:
ΔH = ΣnΔHf(products) - ΣnΔHf(reactants)
where ΔH is the enthalpy change for the reaction, ΣnΔHf(products) is the sum of the enthalpies of formation of the products, and ΣnΔHf(reactants) is the sum of the enthalpies of formation of the reactants.
In this case, the reactants are Mg and HCl, and the products are MgCl2 and H2. The heat of formation values you provided are as follows:
Hf(HCl) = -118.53 kJ/mol
Hf(MgCl2) = -774 kJ/mol
Now, let's calculate the enthalpy change for the reaction:
ΔH = [nHf(MgCl2) + nHf(H2)] - [nHf(Mg) + nHf(HCl)]
In the equation above, n is the stoichiometric coefficient of each substance in the balanced chemical equation.
The balanced chemical equation you provided is: Mg + 2HCl -> MgCl2 + H2
Comparing the coefficients in the equation to the stoichiometric coefficients, we find that:
n(Mg) = 1
n(HCl) = 2
n(MgCl2) = 1
n(H2) = 1
Now we can substitute these values into the equation to calculate the enthalpy change:
ΔH = [1 * -774 kJ/mol + 1 * 0 kJ/mol] - [1 * 0 kJ/mol + 2 * -118.53 kJ/mol]
Simplifying the equation:
ΔH = -774 kJ/mol - 2 * -118.53 kJ/mol
ΔH = -774 kJ/mol + 237.06 kJ/mol
ΔH = -536.94 kJ/mol
Therefore, the actual value for the heat of reaction based on the enthalpies of formation is -536.94 kJ/mol.