The Mistake Bank
Chapter 6: Thermodynamics
Physics says one thing, Chemistry says another. Don't mix them up!
The Sign Convention War
Work DoneScenario: A gas expands from volume $V_1$ to $V_2$ against external pressure $P$.
Student uses the Physics formula:
$$ W = P \Delta V $$
(In Chemistry, this gives the wrong sign!)
Chemistry Focuses on the System!
Work done BY the system (expansion) represents energy leaving the system.
$$ W = - P_{ext} \Delta V $$
- Compression (+ve work)
- Expansion (-ve work)
The Entropy Unit Trap
Gibbs EnergyScenario: Calculate $\Delta G$ if $\Delta H = -40 \text{ kJ}$ and $\Delta S = -20 \text{ J/K}$ at $300 \text{ K}$.
Student plugs values directly:
$$ \Delta G = -40 - (300 \times -20) $$
$$ \Delta G = -40 + 6000 = +5960 $$
(Disaster! You mixed kJ and J.)
Convert Units First!
$\Delta S$ is almost always in Joules (J).
$\Delta H$ is almost always in Kilojoules (kJ).
$$ \Delta S = -20 \text{ J/K} = -0.020 \text{ kJ/K} $$
$$ \Delta G = -40 - (300)(-0.020) $$
$$ = -40 + 6 = \mathbf{-34 \text{ kJ}} $$
Isothermal vs. Adiabatic
Internal EnergyScenario: An ideal gas expands adiabatically ($q=0$). Does temperature change?
Student assumes "Expansion means constant temperature" (confusing it with Isothermal).
Or assumes since $q=0$, $T$ must be constant.
Work comes from Internal Energy!
Since heat ($q$) cannot enter to compensate for the work done ($w$), the energy comes from the gas itself ($\Delta U$).
$$ \Delta U = w $$
Since the gas loses energy doing work, $\Delta U$ is negative, so Temperature Drops (Cooling).
Free Expansion
Work DoneScenario: Calculate work done when a gas expands into a vacuum (Free Expansion).
Student uses the log formula:
$$ w = -2.303 nRT \log(V_2/V_1) $$
(This formula assumes a reversible process against pressure.)
No Opposing Force = No Work!
In a vacuum, external pressure $P_{ext} = 0$.
$$ w = - P_{ext} \Delta V $$
$$ w = 0 \times \Delta V = \mathbf{0} $$
Also, for ideal gas free expansion, $q=0$ and $\Delta U=0$.
Delta H vs. Delta U
ThermochemistryScenario: For the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$, calculate $\Delta H - \Delta U$ at temperature T.
Student forgets to count the moles of gas correctly or ignores the formula.
Assumes $\Delta H = \Delta U$.
Use $\Delta n_g RT$
$$ \Delta H = \Delta U + \Delta n_g RT $$
$$ \Delta n_g = (\text{moles gaseous products}) - (\text{moles gaseous reactants}) $$
$$ \Delta n_g = 2 - (1 + 3) = -2 $$
Answer: $\mathbf{-2RT}$
Spontaneity Criteria
Second LawScenario: For a reaction to be spontaneous, what must be positive?
Student answers: "$\Delta S_{system}$ must be positive."
(Wrong! Systems can become ordered spontaneously, e.g., freezing water below 0°C.)
Total Entropy matters!
$$ \Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0 $$
OR, in terms of Gibbs Energy (system only):
$$ \Delta G_{sys} < 0 $$
Always check $\Delta G$, not just $\Delta S_{sys}$.
Confess Your Sins!
"Thermodynamics is a funny subject. The first time you go through it, you don't understand it at all." - Arnold Sommerfeld
Did one of these catch you? Or do you have a different horror story from your last exam?
Scroll down to the comments section below and tell us:
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