Predict the sign of ( ∆H) in each of the following reactions:
NH3 (g) + HCl (g) NH4Cl(s) ∆S = +, ∆G = -
Fe (s) + Cu+2 (aq) Fe+2 (aq) + Cu(s) ∆S = +, ∆G = -
Cl2 (g) + 2Na (s) 2NaCl (s) ∆S = +, ∆G = -
To predict the sign of ∆H (the enthalpy change) in each of the given reactions, we need to consider the values of ∆S (entropy change) and ∆G (free energy change).
In general, when ∆G is negative (∆G < 0), it indicates that the reaction is spontaneous in the forward direction. This suggests that the products are more stable than the reactants and that the reaction will release energy (exothermic). On the other hand, when ∆G is positive (∆G > 0), the reaction is non-spontaneous in the forward direction, indicating that the reactants are more stable than the products and that the reaction will require input of energy (endothermic).
Now, let's analyze each reaction:
1. NH3 (g) + HCl (g) → NH4Cl (s):
- ∆S = positive (+) indicates an increase in randomness (entropy) going from gaseous reactants to a solid product.
- ∆G = negative (-) suggests that the reaction is spontaneous and proceeds in the forward direction.
Thus, the reaction is exothermic (∆H < 0), as energy is released.
2. Fe (s) + Cu+2 (aq) → Fe+2 (aq) + Cu (s):
- ∆S = positive (+) implies an increase in entropy when going from a solid to a solution and also forming a solid.
- ∆G = negative (-) indicates a spontaneous reaction in the forward direction.
Therefore, the reaction is also exothermic (∆H < 0) since energy is released.
3. Cl2 (g) + 2Na (s) → 2NaCl (s):
- ∆S = positive (+) suggests an increase in entropy when going from a gas to a solid.
- ∆G = negative (-) implies a spontaneous reaction in the forward direction.
Thus, this reaction is exothermic (∆H < 0) as energy is released.
In conclusion, for all the given reactions, the sign of ∆H is negative (∆H < 0), indicating an exothermic process where energy is released.