Online Test — Equilibrium

18 Questions • 15 min • Chapter MCQ

15:00

Question 1 of 18

At chemical equilibrium:

the reaction has stopped

forward and reverse rates are equal

only the forward reaction occurs

concentrations keep changing

Explanation: Equilibrium is dynamic — forward and reverse rates are equal, so concentrations stay constant.

Question 2 of 18

For $N_2+3H_2\rightleftharpoons 2NH_3$, $K_p=K_c(RT)^{\Delta n}$ with $\Delta n=$

$+2$

$0$

$-2$

$-1$

Explanation: $\Delta n=2-(1+3)=-2$.

Question 3 of 18

If $Q_c

reverse

forward

neither

stops

Explanation: $Q

Question 4 of 18

Only this factor changes the value of $K$:

concentration

pressure

catalyst

temperature

Explanation: $K$ depends only on temperature.

Question 5 of 18

A catalyst added to an equilibrium system:

shifts it forward

shifts it reverse

does not shift it

increases $K$

Explanation: It speeds both directions equally; the position and $K$ are unchanged.

Question 6 of 18

For $H_2+I_2\rightleftharpoons 2HI$ with $[H_2]=[I_2]=1\ \text{M}$ and $[HI]=4\ \text{M}$, $K_c$ is:

Explanation: $K_c=\dfrac{4^2}{1\times1}=16$.

Question 7 of 18

A Brønsted–Lowry acid is a:

proton acceptor

proton donor

electron-pair donor

electron-pair acceptor

Explanation: A Brønsted–Lowry acid donates a proton.

Question 8 of 18

The conjugate base of $H_2SO_4$ is:

$SO_4^{2-}$

$HSO_4^-$

$H_3SO_4^+$

$SO_3$

Explanation: Removing one $H^+$ from $H_2SO_4$ gives $HSO_4^-$.

Question 9 of 18

A Lewis base is a species that:

accepts a proton

donates a proton

donates an electron pair

accepts an electron pair

Explanation: A Lewis base donates an electron pair.

Question 10 of 18

The pH of $0.001\ \text{M}$ HCl is:

Explanation: $[H^+]=10^{-3}$, so $pH=3$.

Question 11 of 18

At $298\ \text{K}$, $K_w=[H^+][OH^-]$ equals:

$10^{-7}$

$10^{-14}$

$10^{-10}$

$1$

Explanation: $K_w=1.0\times10^{-14}$ at $298\ \text{K}$.

Question 12 of 18

For a $0.1\ \text{M}$ weak acid with $K_a=10^{-5}$, $[H^+]$ is about:

$10^{-3}\ \text{M}$

$10^{-5}\ \text{M}$

$10^{-1}\ \text{M}$

$10^{-6}\ \text{M}$

Explanation: $[H^+]=\sqrt{K_a c}=\sqrt{10^{-5}\times0.1}=\sqrt{10^{-6}}=10^{-3}\ \text{M}$.

Question 13 of 18

An aqueous solution of $CH_3COONa$ is:

acidic

neutral

basic

amphoteric

Explanation: Salt of a weak acid and strong base; the acetate ion hydrolyses to give a basic solution.

Question 14 of 18

The Henderson–Hasselbalch equation for an acidic buffer is:

$pH=pK_a-\log\dfrac{[salt]}{[acid]}$

$pH=pK_a+\log\dfrac{[acid]}{[salt]}$

$pH=pK_a+\log\dfrac{[salt]}{[acid]}$

$pH=pK_b+\log\dfrac{[salt]}{[base]}$

Explanation: $pH=pK_a+\log\dfrac{[salt]}{[acid]}$.

Question 15 of 18

A buffer with $[salt]=[acid]$ has pH equal to:

$7$

$pK_a$

$0$

$14$

Explanation: $\log 1=0$, so $pH=pK_a$.

Question 16 of 18

For a salt $AB_2$ of solubility $s$, $K_{sp}$ is:

$s^2$

$4s^3$

$27s^4$

$2s^2$

Explanation: $[A^{2+}]=s$, $[B^-]=2s$, so $K_{sp}=s(2s)^2=4s^3$.

Question 17 of 18

Precipitation of a sparingly soluble salt occurs when:

$Q_{sp}

$Q_{sp}=K_{sp}$

$Q_{sp}>K_{sp}$

$K_{sp}=0$

Explanation: When the ionic product exceeds $K_{sp}$, the salt precipitates.

Question 18 of 18

Adding $NaCl$ to a saturated solution of $AgCl$ will:

increase its solubility

decrease its solubility

increase $K_{sp}$

have no effect

Explanation: The common $Cl^-$ ion lowers $[Ag^+]$ to keep $K_{sp}$ constant, so $AgCl$ solubility decreases.

0 / 18

Test Complete!

Score

Correct

Wrong

0:00

Time

Back to Study All Chapters