20) Identify the location of reduction in an electrochemical cell. A) the socket B) the salt bridge C) the anode D) the electrode E) the cathode 21) The value of AG° at 141.0°C for the formation of phosphorous trichloride from its constituent elements, P2 (g) + 3Cl2 (g) 2PC13 (g) is kJ/mol. At 25.0°C for this reaction, AH is -720.5 kJ/mol, AG° is -642.9 kJ/mol, and AS is -263.7 J/K. A) -829.7 D) 1.08 × 105 B) 3.65 x 104 22) Phosphorous and chlorine gases combine to produce phosphorous trichloride: P2 (g) + 3Cl2 (g) 2PC13 (g) B) -3.88 × 103 X C) -7.28 × 103 D) -708.4 E) -44.2 C) -612.3 ?>> AGO at 298 K for this reaction is -642.9 kJ/mol. The value of AG at 298 K for a reaction mixture that consists of 1.5 atm P2, 1.6 atm Cl2, and 0.65 atm PCl3 is A) -649.5 E) -683.3
The Correct Answer and Explanation is :
Let’s go through each of the questions systematically:
20) Identify the location of reduction in an electrochemical cell.
The reduction process in an electrochemical cell occurs at the cathode. The cathode is where electrons are gained by the species involved in the reduction reaction. Therefore, the correct answer is:
E) the cathode
Explanation:
In electrochemical cells, oxidation happens at the anode (where electrons are lost), and reduction occurs at the cathode (where electrons are gained). This is a fundamental concept of redox reactions.
21) The value of ΔG° at 141.0°C for the formation of phosphorus trichloride from its constituent elements, P₂ (g) + 3Cl₂ (g) → 2PCl₃ (g) is kJ/mol.
We are given the following data:
- ΔH° = -720.5 kJ/mol at 25.0°C
- ΔG° = -642.9 kJ/mol at 25.0°C
- ΔS° = -263.7 J/K at 25.0°C
We need to calculate the standard Gibbs free energy (ΔG°) at 141.0°C (which is 414.15 K).
The formula to calculate ΔG° is:
[
\Delta G^\circ = \Delta H^\circ – T \Delta S^\circ
]
First, convert ΔS° into kJ/K (because ΔH° is in kJ/mol):
[
\Delta S^\circ = -263.7 \, \text{J/K} = -0.2637 \, \text{kJ/K}
]
Now, substitute the values into the formula:
[
\Delta G^\circ = -720.5 – (414.15 \times -0.2637)
]
Calculating this:
[
\Delta G^\circ = -720.5 + 109.4 = -611.1 \, \text{kJ/mol}
]
However, the closest value is not directly calculated in this way, and typically, we’d need to apply more specific temperature dependencies or data. But based on typical answer choices, the closest approximation can be:
C) -612.3
22) Given the reaction: P₂ (g) + 3Cl₂ (g) → 2PCl₃ (g), and we know that at 298 K, ΔG° = -642.9 kJ/mol, we need to calculate ΔG for a mixture with partial pressures:
- P₂ = 1.5 atm
- Cl₂ = 1.6 atm
- PCl₃ = 0.65 atm
To calculate ΔG under non-standard conditions, we use the formula:
[
\Delta G = \Delta G^\circ + RT \ln Q
]
Where:
- ( \Delta G^\circ = -642.9 \, \text{kJ/mol} )
- ( R = 8.314 \times 10^{-3} \, \text{kJ/mol·K} ) (gas constant)
- ( T = 298 \, \text{K} )
- ( Q ) is the reaction quotient, given by:
[
Q = \frac{{[PCl_3]^2}}{{[P_2] \cdot [Cl_2]^3}}
]
Substitute the given values:
[
Q = \frac{{(0.65)^2}}{{(1.5) \cdot (1.6)^3}}
]
First, calculate Q:
[
Q = \frac{{0.4225}}{{1.5 \cdot 4.096}} = \frac{{0.4225}}{{6.144}} = 0.0688
]
Now, use this to calculate ΔG:
[
\Delta G = -642.9 + (8.314 \times 10^{-3} \times 298 \times \ln 0.0688)
]
[
\Delta G = -642.9 + (2.478 \times \ln 0.0688)
]
Since ( \ln 0.0688 \approx -2.673 ):
[
\Delta G = -642.9 + (2.478 \times -2.673) = -642.9 – 6.62 = -649.5 \, \text{kJ/mol}
]
So, the correct answer is:
A) -649.5
Final Answer Summary:
- 20) E) the cathode
- 21) C) -612.3 kJ/mol
- 22) A) -649.5 kJ/mol