The normal freezing point of ammonia is 78 degrees celsius.Predict the signs of ?H, and ?G for ammonia when itfreezes at 0 degrees celsius and 1 atm: NH3(l)–>NH3(s)
**I KNOW ITS NOT A
?H ?S ?G
A – – 0
B – + –
C + – +
D + + 0
E – – –
The correct answer and explanation is:
To predict the signs of ΔH\Delta H, ΔS\Delta S, and ΔG\Delta G for the freezing of ammonia (NH3(l)→NH3(s)NH_3(l) \rightarrow NH_3(s)) at 0∘C0^\circ C and 11 atm, let’s analyze the thermodynamics of the process.
Correct Answer: B – ΔH:−\Delta H: -, ΔS:+\Delta S: +, ΔG:−\Delta G: –
Explanation:
- ΔH\Delta H (Change in Enthalpy): Negative (-)
Freezing is an exothermic process because energy is released when ammonia molecules transition from the liquid phase to the solid phase. During freezing, the intermolecular forces (e.g., hydrogen bonds in ammonia) are strengthened, releasing heat. Thus, ΔH\Delta H is negative. - ΔS\Delta S (Change in Entropy): Positive (+)
Entropy (SS) is a measure of disorder or randomness in a system. Freezing involves a phase change from liquid (more disordered) to solid (more ordered). However, **at 0∘C0^\circ C, ammonia is freezing at a temperature much lower than its normal freezing point (78∘C78^\circ C), so it freezes non-spontaneously, requiring an increase external entropy}to.
Apologies for the confusion earlier! Let me correctly evaluate the situation and provide a clear explanation.
The freezing of ammonia (NH3(l)→NH3(s)NH_3(l) \to NH_3(s)) at 0∘C0^\circ C and 1 atm1 \, \text{atm} involves the following thermodynamic considerations:
Correct Answer: E – ΔH:−\Delta H: -, ΔS:−\Delta S: -, ΔG:−\Delta G: –
Detailed Explanation:
- ΔH\Delta H (Change in Enthalpy): Negative (-)
Freezing is an exothermic process because energy is released when the liquid ammonia molecules transition to a more stable, solid structure. This release of heat to the surroundings indicates that ΔH\Delta H is negative. - ΔS\Delta S (Change in Entropy): Negative (-)
Entropy (SS) measures disorder. When ammonia freezes, the molecules go from a more disordered liquid state to a highly ordered solid crystalline structure. This reduction in disorder means ΔS\Delta S is negative. - ΔG\Delta G (Change in Gibbs Free Energy): Negative (-)
The Gibbs free energy (ΔG=ΔH−TΔS\Delta G = \Delta H – T\Delta S) determines spontaneity. At 0∘C0^\circ C, which is significantly below ammonia’s normal freezing point (78∘C78^\circ C), freezing occurs spontaneously. A negative ΔH\Delta H and the large magnitude of the TΔST\Delta S term (negative due to ΔS\Delta S) combine to make ΔG\Delta G negative.
Key Notes:
- At 78∘C78^\circ C, the freezing/melting process would be at equilibrium (ΔG=0\Delta G = 0).
- At 0∘C0^\circ C, freezing is thermodynamically favorable (ΔG<0\Delta G < 0) because ammonia is far below its normal freezing point, driving the phase transition to solid despite the accompanying decrease in entropy.