A scientist measures the standard enthalpy change for the following reaction to be -53.4 kJ :
Ca(OH)2(aq) + 2 HCl(aq) CaCl2(s) + 2 H2O(l)
Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O(l) is kJ/mol
Do you have the dHf values for the other materials in the reaction? I don't have them in my text.
dHrxn = (n*dHf products) - (n*dH reactants). Substitute and solve for dH H2O.
To determine the standard enthalpy of formation of H2O(l), we need to use the given information and apply Hess's Law. Hess's Law states that the overall enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states.
First, we need to write down the balanced equation for the reaction:
Ca(OH)2(aq) + 2 HCl(aq) → CaCl2(s) + 2 H2O(l)
Next, we can calculate the standard enthalpy change (∆H) for this reaction using the given value of -53.4 kJ. The ∆H of this reaction represents the sum of the standard enthalpies of formation (∆Hf) of the products minus the sum of the standard enthalpies of formation of the reactants.
∆H = ∑∆Hf(products) - ∑∆Hf(reactants)
Using the values provided in standard enthalpies of formation table, we can substitute the values and solve for the unknown ∆Hf of H2O(l).
Let's write the balanced equation in terms of the enthalpies of formation:
Ca(OH)2(aq) → Ca(OH)2(aq) (∆Hf = 0 kJ/mol)
2 HCl(aq) → 2 HCl(aq) (∆Hf = 0 kJ/mol)
CaCl2(s) → CaCl2(s) (∆Hf = 0 kJ/mol)
2 H2O(l) → 2 H2O(l) (∆Hf = ? kJ/mol)
Now, we can rearrange the equation:
-53.4 kJ/mol = 0 kJ/mol + 0 kJ/mol + 0 kJ/mol - 2(∆Hf)
Simplifying, we have:
-53.4 kJ/mol = -2(∆Hf)
Dividing both sides by -2:
∆Hf = 26.7 kJ/mol
Therefore, the standard enthalpy of formation of H2O(l) is 26.7 kJ/mol.