IMO Practice Test — Thermodynamics
14 Questions • 15 min • Olympiad level
15:00
Question 1 of 14
A gas absorbs 500 J and does 200 J of work. $\Delta U$ is:
$+700$ J
$+300$ J
$-300$ J
$+200$ J
Explanation: $w=-200$ J; $\Delta U=500-200=300$ J.
Question 2 of 14
For $2\text{SO}_2+\text{O}_2\rightarrow 2\text{SO}_3$, $\Delta n_g$ is:
$0$
$-1$
$+1$
$-2$
Explanation: Products 2, reactants 3; $\Delta n_g=2-3=-1$.
Question 3 of 14
2 mol of an ideal gas expands isothermally and reversibly to twice its volume at 300 K. The work done by the gas is (R=8.314):
$3.46$ kJ
$1.73$ kJ
$5.76$ kJ
$0.69$ kJ
Explanation: $w=nRT\ln 2=2(8.314)(300)(0.693)=3458\ \text{J}\approx3.46$ kJ.
Question 4 of 14
Given $\Delta_c H$: C=$-393.5$, H₂=$-285.8$, C₂H₆(combustion of ethane uses $\Delta_f$)... If $\Delta_f H(\text{CO}_2)=-393.5$ and $\Delta_f H(\text{H}_2\text{O})=-285.8$, then $\Delta_c H$ of $\text{H}_2$ is:
$-393.5$ kJ
$-285.8$ kJ
$-571.6$ kJ
$-110.5$ kJ
Explanation: Combustion of H₂ gives 1 mol H₂O(l), so $\Delta_c H=-285.8$ kJ/mol.
Question 5 of 14
Estimate $\Delta H$ for $\text{CH}_4\rightarrow\text{C}+4\text{H}$ if each C–H bond is 416 kJ/mol:
$+416$ kJ
$+1664$ kJ
$-1664$ kJ
$+832$ kJ
Explanation: Breaking 4 C–H bonds: $4\times416=1664$ kJ absorbed.
Question 6 of 14
For NaCl Born–Haber: if $\Delta_f H=-411$, $\Delta_{sub}=108$, ½D=121, IE=496, $\Delta_{eg}=-349$, lattice enthalpy is:
$-787$ kJ
$+787$ kJ
$-376$ kJ
$-411$ kJ
Explanation: $\Delta_{lattice}=\Delta_f H-(\Delta_{sub}+\tfrac12 D+IE+\Delta_{eg})=-411-376=-787$ kJ.
Question 7 of 14
A reaction has $\Delta H=-50$ kJ, $\Delta S=-100\ \text{J/K}$. At what temperature is $\Delta G=0$?
$200$ K
$500$ K
$250$ K
$1000$ K
Explanation: $T=\Delta H/\Delta S=50000/100=500$ K.
Question 8 of 14
If $\Delta G^\circ=-RT\ln K$ and $K=1$, then $\Delta G^\circ$ is:
negative
positive
zero
infinite
Explanation: $\ln 1=0$, so $\Delta G^\circ=0$.
Question 9 of 14
Vaporisation of 1 mol water at 373 K with $\Delta H=40.7$ kJ gives $\Delta S$ of:
$+109\ \text{J/K}$
$-109\ \text{J/K}$
$+40.7\ \text{J/K}$
$+373\ \text{J/K}$
Explanation: At equilibrium $\Delta S=\Delta H/T=40700/373=109\ \text{J/K}$.
Question 10 of 14
For an ideal monatomic gas $C_v=\tfrac32 R$. The ratio $\gamma=C_p/C_v$ is:
$1.40$
$1.33$
$1.67$
$2.00$
Explanation: $C_p=\tfrac52 R$, so $\gamma=5/3=1.67$.
Question 11 of 14
When the number of gaseous molecules increases in a reaction, $\Delta S$ is generally:
negative
zero
positive
undefined
Explanation: More gas molecules means greater disorder, so $\Delta S>0$.
Question 12 of 14
A reaction with $\Delta H>0$ and $\Delta S>0$ is spontaneous:
at all $T$
never
only at high $T$
only at low $T$
Explanation: The $-T\Delta S$ term overtakes $\Delta H$ when $T$ is large.
Question 13 of 14
Heat released when 1 mol HCl forms from H–H=435, Cl–Cl=242, H–Cl=431 (per mol HCl):
$-92.5$ kJ
$-185$ kJ
$+185$ kJ
$-431$ kJ
Explanation: Per 2 HCl: $-185$ kJ, so per mole HCl it is $-92.5$ kJ.
Question 14 of 14
At constant pressure, the heat absorbed equals:
$\Delta U$
$\Delta H$
$\Delta G$
$T\Delta S$
Explanation: $q_P=\Delta H$.