1.Calculate the enthalpy change (ΔHºrxn) for the following reaction: (The enthalpy of formation of aqueous sodium hydroxide is -469.60 kJ/mol. The enthalpy of formation of liquid water is -285.8 kJ/mol.)
2Na (s) + 2H2O (l) → 2NaOH (aq) + H2 (g)
2.Do you know if the reaction is endothermic, thermodynamic, exothermic, or in its standard state?
3. The surroundings of the reaction will either
-get hotter,
-get cooler,
-neither get hotter nor get cooler due to the 2nd Law of Thermodynamics,
-get cooler if ΔG is negative; get hotter if ΔG is positive
1. DHr = delta Hf
DHfrxn = (n*DHfproducts) - (n*DHfreactants)
From the number you obtain in the above you will have the information to answer 2 and 3.
To calculate the enthalpy change (ΔHºrxn) for the given reaction, you can use the concept of Hess's Law. Hess's Law states that the enthalpy change of a reaction depends only on the initial and final states of the reaction, not on the pathway taken.
1. Start by writing the balanced equation for the reaction:
2Na (s) + 2H2O (l) → 2NaOH (aq) + H2 (g)
2. Determine the enthalpy change for each individual formation reaction:
a) NaOH (aq): ΔHºf = -469.60 kJ/mol
b) H2O (l): ΔHºf = -285.8 kJ/mol
3. Use the stoichiometric coefficients of the balanced equation to determine the overall enthalpy change:
ΔHºrxn = (2 mol of NaOH) * ΔHºf of NaOH + (1 mol of H2O) * ΔHºf of H2O
ΔHºrxn = (2 mol) * (-469.60 kJ/mol) + (1 mol) * (-285.8 kJ/mol)
Simplifying the equation, we get:
ΔHºrxn = -939.20 kJ - 285.8 kJ
ΔHºrxn = -1225.00 kJ
Therefore, the enthalpy change (ΔHºrxn) for this reaction is -1225.00 kJ.
Now, let's proceed to the next questions:
2. To determine if the reaction is endothermic or exothermic in its standard state, you can check the sign of the enthalpy change (ΔHºrxn). If the ΔHºrxn is negative, it indicates that the reaction is exothermic, meaning it releases energy in the form of heat. In this case, since ΔHºrxn is -1225.00 kJ (negative), the reaction is exothermic.
3. According to the Second Law of Thermodynamics, the surroundings of the reaction will either get hotter or cooler depending on the sign of the Gibbs free energy change (ΔG). If ΔG is negative, the reaction is thermodynamically favorable, and the surroundings will get cooler. On the other hand, if ΔG is positive, the reaction is thermodynamically unfavorable, and the surroundings will get hotter.