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 ΔH (the enthalpy change) in each of the given reactions, we need to consider the ΔS (entropy change) and ΔG (Gibbs free energy change).
For the first reaction, NH3 (g) + HCl (g) → NH4Cl(s), we are told that ΔS is positive and ΔG is negative. A positive ΔS indicates an increase in disorder or randomness in the system, while a negative ΔG suggests that the reaction is spontaneous or energetically favorable. Based on these values, we can infer that the reaction is exothermic, meaning ΔH is negative.
For the second reaction, Fe (s) + Cu+2 (aq) → Fe+2 (aq) + Cu(s), we are given that ΔS is positive and ΔG is negative. Again, a positive ΔS indicates an increase in disorder, and a negative ΔG suggests the reaction is spontaneous. Therefore, we can conclude that the reaction is also exothermic, implying a negative ΔH.
Lastly, for the third reaction, Cl2 (g) + 2Na (s) → 2NaCl (s), we are again told that ΔS is positive and ΔG is negative. Following the pattern we discussed earlier, we can conclude that the reaction is exothermic and ΔH is negative.
In summary, based on the provided information, the sign of ΔH is negative (exothermic) for all three reactions.