Each of the following processes is spontaneous as described. Which one will have ∆H approximately equal to zero?
1. precipitating 1 mol of AgBr(s) from a
solution of Ag+(aq) and Br−(aq)
2. 1 L He(1 atm, 298 K) + 1 L Ar(1 atm,
298 K) → 2 L He/Ar mixture(1 atm, 298 K)
3. freezing 1 mol of water at − 10◦C
4. evaporating 1 mol of water at 100◦C and 1 atm
2. 1 L He(1 atm, 298 K) + 1 L Ar(1 atm, 298 K) → 2 L He/Ar mixture(1 atm, 298 K)
To determine which process will have ΔH approximately equal to zero, we need to look at the enthalpy change (ΔH) values for each process.
1. Precipitating 1 mol of AgBr(s) from a solution of Ag+(aq) and Br−(aq):
When a precipitate forms, energy is released, resulting in a negative ΔH. Therefore, this process will have ΔH less than zero.
2. 1 L He(1 atm, 298 K) + 1 L Ar(1 atm, 298 K) → 2 L He/Ar mixture(1 atm, 298 K):
Mixing different gases does not involve any significant energy changes, so the ΔH for this process is also expected to be close to zero.
3. Freezing 1 mol of water at -10°C:
Freezing is an exothermic process, meaning it releases energy. Water freezing at a low temperature will have a negative ΔH.
4. Evaporating 1 mol of water at 100°C and 1 atm:
Vaporizing or evaporating water requires energy input, making it an endothermic process. ΔH for this process is expected to be positive.
Therefore, out of the given processes, the process with ΔH approximately equal to zero is the second process: 1 L He(1 atm, 298 K) + 1 L Ar(1 atm, 298 K) → 2 L He/Ar mixture(1 atm, 298 K).
To determine which process will have ∆H approximately equal to zero, we need to consider the enthalpy changes associated with each process.
1. Precipitating 1 mol of AgBr(s) from a solution of Ag+(aq) and Br−(aq): This process involves the formation of a solid from aqueous ions. Since AgBr(s) is a solid, the enthalpy change (∆H) for this process is likely to be negative since the solid formed is more stable than the aqueous ions.
2. 1 L He(1 atm, 298 K) + 1 L Ar(1 atm, 298 K) → 2 L He/Ar mixture(1 atm, 298 K): This process involves mixing two ideal gases. Since there is no change in bonding or phase, the enthalpy change (∆H) for this process is expected to be close to zero. Ideal gases follow the ideal gas law and intermolecular forces are negligible in this case.
3. Freezing 1 mol of water at -10◦C: This process involves the conversion of liquid water to solid ice. The enthalpy change (∆H) for this process is negative because heat is released as the water molecules come together to form a more ordered crystal lattice.
4. Evaporating 1 mol of water at 100◦C and 1 atm: This process involves the conversion of liquid water to gaseous water vapor. The enthalpy change (∆H) for this process is positive because energy is needed to break the intermolecular forces and convert the water molecules into a less ordered gas phase.
Based on these explanations, the process that will have ∆H approximately equal to zero is process 2, which is the mixing of helium and argon gases.