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Question #44191

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The Question

Consider a thermodynamic system comprising a metal lattice \ce{M(s)} and its bound electrons at $T = 300\text{ K}$ and $P = 1\text{ bar}$ . The system is irradiated with $1\text{ mole}$ of monochromatic photons having a total energy of $E_{\text{photon}} = 450\text{ kJ mol}^{-1}$ , triggering the photoelectric emission of $1\text{ mole}$ of electrons. The molar work function of the metal is $\Phi = 300\text{ kJ mol}^{-1}$ , which strictly corresponds to the standard enthalpy change ( $\Delta H^\circ$ ) for the emission of electrons to a vacuum at rest. The standard entropy change for this specific state transition is $\Delta S^\circ = +100\text{ J K}^{-1}\text{ mol}^{-1}$ .

If the final thermodynamic state of the system is defined strictly as the metal cation lattice \ce{M+(s)} and the emitted electrons at rest in a vacuum, what is the standard Gibbs free energy change ( $\Delta G^\circ$ ) for this state transition, and how is the maximum kinetic energy ( $KE_{\text{max}}$ ) of the emitted electrons classified thermodynamically?

Options

A.
$\Delta G^\circ = +270\text{ kJ mol}^{-1}$ ; $KE_{\text{max}}$ is a path-dependent quantity determined by the incident photon frequency.
B.
$\Delta G^\circ = +420\text{ kJ mol}^{-1}$ ; $KE_{\text{max}}$ is a state function representing the internal energy of the emitted electrons.
C.
$\Delta G^\circ = -150\text{ kJ mol}^{-1}$ ; $KE_{\text{max}}$ acts as a thermodynamic driving force, making the state transition spontaneous.
D.
$\Delta G^\circ = +150\text{ kJ mol}^{-1}$ ; $KE_{\text{max}}$ is a path-independent constant for a given incident radiation.

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Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Photoelectric Effect Threshold|Spontaneity Criteria|State vs Path Functions

Why This Question Is Tricky

Common Mistakes

This problem tests the critical distinction between State Functions and Path Functions within the context of quantum mechanical phenomena (Photoelectric Effect) and macroscopic thermodynamics.

Students often fall into the trap of using the incident photon energy ( $450\text{ kJ mol}^{-1}$ ) as the enthalpy change ( $\Delta H$ ), leading to an incorrect $\Delta G^\circ$ of $+420\text{ kJ mol}^{-1}$ (Option B).

Alternatively, they may subtract the work function from the photon energy and assume the resulting kinetic energy drives spontaneity, yielding a negative $\Delta G^\circ$ (Option C).

The core exception being tested is that while the photon delivers energy (a path-dependent work/heat input), the thermodynamic state of the system is defined strictly by the potential energy difference (Work Function) and entropy difference between the initial lattice and the final 'at rest' state.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Structure of Atom - Class 11 Part 1
Concept	Photoelectric Effect Threshold
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free