Calculate Delta H for the following equation: Zn(s) + 2H+(aq) = Zn2+(aq) + H2(g)
dHrxb = (n*dHformation products)-(n*dHformation reactants)
To calculate ΔH for the given equation, we need to use Hess's Law or known standard enthalpies of formation.
Hess's Law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of its individual steps. In this case, we'll break the reaction into two steps: the formation of Zn2+(aq) from Zn(s) and the formation of H2(g) from 2H+(aq).
Step 1: Zn(s) → Zn2+(aq)
To find the enthalpy change for this step, we need to know the standard enthalpy of formation (ΔHf°) of Zn(s) and Zn2+(aq). ΔHf° represents the enthalpy change when one mole of a compound is formed from its elements in their standard states at 25°C and 1 atm pressure.
Assuming that Zn(s) is in its standard state, its ΔHf° is 0 kJ/mol. The ΔHf° value for Zn2+(aq) is -153.9 kJ/mol.
Step 2: 2H+(aq) → H2(g)
Again, we need to know the ΔHf° values of H+(aq) and H2(g). The ΔHf° of H+(aq) is 0 kJ/mol because it is the standard state for hydrogen ions. The ΔHf° of H2(g) is 0 kJ/mol because it is also in the standard state.
Now, we can calculate the ΔH for the overall reaction using the enthalpy changes of the individual steps:
ΔH = ΔHf°(products) - ΔHf°(reactants)
= [ΔHf°(Zn2+(aq)) + ΔHf°(H2(g))] - [ΔHf°(Zn(s)) + ΔHf°(2H+(aq))]
= [-153.9 kJ/mol + 0 kJ/mol] - [0 kJ/mol + 0 kJ/mol]
= -153.9 kJ/mol
Therefore, the enthalpy change (ΔH) for the given equation is -153.9 kJ/mol.