IB Chemistry SL Reactivity 2: How much, how fast and how far? questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Type

Multiple choice Written

Difficulty

Easy Medium Hard

Question 1

Multiple choiceHard1 mark

Magnesium reacts with hydrochloric acid according to the equation below. The hydrogen is collected at the conditions for which the molar gas volume is $24.0\ \mathrm{dm^3\,mol^{-1}}$ Use $A_\mathrm{r}(\mathrm{Mg})=24.31$

\mathrm{Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)}

Substance	Amount
Mg	$0.120\ \mathrm{g}$
HCl(aq)	$25.0\ \mathrm{cm^3}$ of $0.200\ \mathrm{mol\,dm^{-3}}$
$H_2$ collected	$56.4\ \mathrm{cm^3}$

Which option gives the limiting reactant and the percentage yield of $H_2$

Show answer and explanation

Answer: C. Compare the available amounts using the $1:2$ mole ratio of Mg to HCl, then calculate the yield relative to the theoretical $H_2$ volume.

$n(\mathrm{Mg})=0.120/24.31=4.94\times10^{-3}\ \mathrm{mol}$ $n(\mathrm{HCl})=0.0250\times0.200=5.00\times10^{-3}\ \mathrm{mol}$ The Mg would require $9.87\times10^{-3}\ \mathrm{mol}$ HCl, so HCl is limiting.
$n(\mathrm{H_2})=5.00\times10^{-3}/2=2.50\times10^{-3}\ \mathrm{mol}$ giving a theoretical volume of $60.0\ \mathrm{cm^3}$
Percentage yield $=(56.4/60.0)\times100=94.0\%$

A: Compares the raw mole amounts to label Mg limiting, ignoring the equation’s coefficients.
B: Reverses the yield ratio, calculating $60.0/56.4\times100=106\%$
D: Treats Mg as limiting and uses its amount to calculate the theoretical $H_2$ volume, giving $118.5\ \mathrm{cm^3}$ and a $47.6\%$ yield.
Remember: compare available amounts after accounting for stoichiometric coefficients to identify the limiting reactant.

Question 2

Multiple choiceHard1 mark

In a sealed vessel, the concentrations in the reversible system $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)}$ become constant while all three gases remain present. Which account correctly describes the reaction rates at this point and gives the expression for $K_c$

Show answer and explanation

Answer: B. Constant concentrations mean there is no net change, so the forward and backward rates are equal. At dynamic equilibrium, both reactions continue, making these rates non-zero. The equilibrium expression uses each concentration raised to its coefficient: $K_c=\dfrac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}$

A: takes the reciprocal, placing reactants over products.
C: treats equilibrium as static and sets both reaction rates to zero.
D: unequal rates would cause net formation of $\mathrm{NH_3}$ so concentrations would change.

Remember: At equilibrium, concentrations are constant; the rates, not the concentrations, are equal.