Physical Mock 1

Thermodynamics and kinetics questions.

3 questions • Estimated time: 30-40 minutes

How to Use This Mock

Read each question carefully
Attempt your own answer first — spend at least 5 minutes thinking
Only reveal the model answer after you've tried
Compare your reasoning to the model answer

Thermodynamics

At 298K, the standard Gibbs free energy of formation of liquid water is $-237 kJ/mol$ while that of gaseous water is $-229 kJ/mol$ . What does this tell us about the spontaneity of water evaporation at room temperature? Why does water still evaporate from a glass at 25°C?

Model Answer

The more negative $\Delta_f G°$ for liquid water tells us that liquid is the thermodynamically stable form at standard conditions (298K, 1 bar). Evaporation: $H_2O(l) \rightarrow H_2O(g)$ has $\Delta G° = -229 - (-237) = +8 kJ/mol$ .

Since $\Delta G° > 0$ , evaporation is non-spontaneous under standard conditions — meaning water vapour at 1 bar pressure would spontaneously condense.

But water still evaporates because real conditions aren't standard conditions. The Gibbs energy depends on concentration/pressure:

$\Delta G = \Delta G° + RT\ln Q$

When water vapour pressure is below the saturated vapour pressure (low humidity), $Q < K$ and $\Delta G < 0$ , making evaporation spontaneous. The system moves toward equilibrium, not toward pure liquid.

This illustrates a crucial distinction: $\Delta G°$ tells us about standard conditions, but $\Delta G$ (which depends on actual conditions) determines real spontaneity.

Key Concepts:

Gibbs Free EnergyStandard vs Non-standard ConditionsChemical EquilibriumVapour Pressure

Kinetics

A reaction has activation energy of $50 kJ/mol$ . By what factor does the rate constant change when temperature increases from 300K to 310K? (R = 8.314 J/mol·K)

Model Answer

Using the Arrhenius equation: $k = A e^{-E_a/RT}$

The ratio of rate constants is:

\frac{k_2}{k_1} = \frac{e^{-E_a/RT_2}}{e^{-E_a/RT_1}} = e^{\frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})}

Substituting values:

\frac{E_a}{R} = \frac{50000}{8.314} = 6014 K

\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300} - \frac{1}{310} = \frac{310-300}{300 \times 310} = \frac{10}{93000} = 1.075 \times 10^{-4} K^{-1}

\frac{k_2}{k_1} = e^{6014 \times 1.075 \times 10^{-4}} = e^{0.647} = 1.91

The rate constant approximately doubles for a 10°C rise — this is a common rule of thumb in chemistry, and now you can see where it comes from!

Note:

The 'rate doubles per 10°C' rule is an approximation that works for activation energies around 50 kJ/mol near room temperature.

Key Concepts:

Arrhenius EquationActivation EnergyTemperature Dependence of Rate

Entropy

Without looking up values, predict the sign of $\Delta S$ for the following processes and explain: (a) Dissolving NaCl in water, (b) Freezing water, (c) $N_2O_4(g) \rightarrow 2NO_2(g)$

Model Answer

**(a) Dissolving NaCl in water: $\Delta S$ is small, slightly positive**

At first, you might think positive — we're breaking an ordered crystal into dispersed ions. But water molecules become ordered around the ions (hydration shells), decreasing their entropy. These effects partially cancel. The net result is slightly positive because the crystal → dispersed ions effect slightly dominates.

**(b) Freezing water: $\Delta S$ is negative**

Liquid water has molecules in disordered motion. Ice has molecules locked in a crystalline lattice. Going from disorder to order means entropy decreases. This is why freezing is only spontaneous when $T\Delta S < \Delta H$ (i.e., below 0°C where the exothermic nature drives the process).

**(c) $N_2O_4(g) \rightarrow 2NO_2(g)$ : $\Delta S$ is positive**

One mole of gas becomes two moles of gas. More gas molecules = more ways to arrange them = more microstates = higher entropy. For gas-phase reactions, counting moles of gas is often a reliable predictor of entropy change.

Key Concepts:

EntropyDissolutionPhase TransitionsGas-Phase Reactions

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