IMO Practice Test — Equilibrium
12 Questions • 15 min • Olympiad level
15:00
Question 1 of 12
If a reaction is reversed, its new equilibrium constant is:
the same $K$
$1/K$
$2K$
$K^2$
Explanation: Reversing a reaction inverts the constant: $K_{reverse}=1/K_{forward}$.
Question 2 of 12
If all coefficients of a reaction are doubled, the constant becomes:
$2K$
$K/2$
$K^2$
$\sqrt{K}$
Explanation: Multiplying the equation by $n$ raises the constant to the power $n$; doubling gives $K^2$.
Question 3 of 12
For $PCl_5\rightleftharpoons PCl_3+Cl_2$ at total pressure $P$ with degree of dissociation $\alpha$, $K_p=$
$\dfrac{\alpha^2}{1-\alpha}P$
$\dfrac{\alpha^2}{1-\alpha^2}P$
$\dfrac{\alpha}{1-\alpha}P$
$\alpha^2 P$
Explanation: Mole fractions give $K_p=\dfrac{\alpha^2}{1-\alpha^2}P$.
Question 4 of 12
$1\ \text{mol}$ $PCl_5$ in $2\ \text{L}$ is $50\%$ dissociated. $K_c$ is:
$0.125$
$0.25$
$0.5$
$1.0$
Explanation: At eq: $PCl_5=0.5$, $PCl_3=Cl_2=0.5\ \text{mol}$; conc $=0.25,0.25,0.25$; $K_c=\dfrac{0.25\times0.25}{0.25}=0.25\ \text{mol L}^{-1}$.
Question 5 of 12
For an exothermic reaction, increasing temperature:
increases $K$
decreases $K$
does not change $K$
doubles the rate only
Explanation: Heat favours the endothermic (reverse) direction, so $K$ decreases.
Question 6 of 12
The pH of $10^{-8}\ \text{M}$ HCl is closest to:
$8$
$7$
just below $7$
$6$
Explanation: Water's own $H^+$ dominates; total $[H^+]\approx1.05\times10^{-7}$, giving $pH$ just below 7 (about 6.98).
Question 7 of 12
If $K_a$ of an acid is $1.0\times10^{-5}$, its $pK_a$ is:
$5$
$-5$
$9$
$14$
Explanation: $pK_a=-\log K_a=-\log(10^{-5})=5$.
Question 8 of 12
The degree of ionisation of a $0.01\ \text{M}$ acid with $K_a=10^{-6}$ is:
$0.001$
$0.01$
$0.1$
$0.0001$
Explanation: $\alpha=\sqrt{K_a/c}=\sqrt{10^{-6}/10^{-2}}=\sqrt{10^{-4}}=10^{-2}=0.01$.
Question 9 of 12
Adding $NH_4Cl$ to aqueous $NH_3$ will:
increase $[OH^-]$
decrease $[OH^-]$
not affect $[OH^-]$
make it strongly acidic
Explanation: Common $NH_4^+$ ion suppresses ionisation of $NH_3$, lowering $[OH^-]$.
Question 10 of 12
A buffer of $0.1\ \text{M}$ acid and $1.0\ \text{M}$ salt ($pK_a=4.7$) has pH:
$3.7$
$4.7$
$5.7$
$6.7$
Explanation: $pH=4.7+\log(1.0/0.1)=4.7+1=5.7$.
Question 11 of 12
$K_{sp}$ of $Ag_2CrO_4$ in terms of solubility $s$ is:
$s^2$
$2s^3$
$4s^3$
$27s^4$
Explanation: $[Ag^+]=2s$, $[CrO_4^{2-}]=s$; $K_{sp}=(2s)^2 s=4s^3$.
Question 12 of 12
Of $AgCl$ ($K_{sp}=1.8\times10^{-10}$) and $AgI$ ($K_{sp}=8.5\times10^{-17}$), the one with higher molar solubility is:
$AgI$
$AgCl$
equal
cannot be decided
Explanation: Both are $AB$ type, so $s=\sqrt{K_{sp}}$; the larger $K_{sp}$ ($AgCl$) is more soluble.