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

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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 dm3 mol−124.0\ \mathrm{dm^3\,mol^{-1}} Use Ar(Mg)=24.31A_\mathrm{r}(\mathrm{Mg})=24.31

Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)\mathrm{Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)}
SubstanceAmount
Mg0.120 g0.120\ \mathrm{g}
HCl(aq)25.0 cm325.0\ \mathrm{cm^3} of 0.200 mol dm−30.200\ \mathrm{mol\,dm^{-3}}
H2H_2 collected56.4 cm356.4\ \mathrm{cm^3}

Which option gives the limiting reactant and the percentage yield of H2H_2

Show answer and explanation

Answer: C. Compare the available amounts using the 1:21:2 mole ratio of Mg to HCl, then calculate the yield relative to the theoretical H2H_2 volume.

  1. n(Mg)=0.120/24.31=4.94×10−3 moln(\mathrm{Mg})=0.120/24.31=4.94\times10^{-3}\ \mathrm{mol} n(HCl)=0.0250×0.200=5.00×10−3 moln(\mathrm{HCl})=0.0250\times0.200=5.00\times10^{-3}\ \mathrm{mol} The Mg would require 9.87×10−3 mol9.87\times10^{-3}\ \mathrm{mol} HCl, so HCl is limiting.
  2. n(H2)=5.00×10−3/2=2.50×10−3 moln(\mathrm{H_2})=5.00\times10^{-3}/2=2.50\times10^{-3}\ \mathrm{mol} giving a theoretical volume of 60.0 cm360.0\ \mathrm{cm^3}
  3. Percentage yield =(56.4/60.0)×100=94.0%=(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 60.0/56.4×100=106%60.0/56.4\times100=106\%
D: Treats Mg as limiting and uses its amount to calculate the theoretical H2H_2 volume, giving 118.5 cm3118.5\ \mathrm{cm^3} and a 47.6%47.6\% yield.
Remember: compare available amounts after accounting for stoichiometric coefficients to identify the limiting reactant.

Question 2

Multiple choiceEasy1 mark

For the reaction shown, Ea=45 kJ mol−1E_\mathrm{a}=45\ \mathrm{kJ\,mol^{-1}} Which proposed collision involves the reactants and has at least this energy?

HH+II2HI
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Answer: D. The forward reaction uses H₂ and I₂ as reactants. The activation energy is the minimum energy colliding particles need for a successful collision, so 45 kJ mol⁻¹ meets the stated threshold.

A: uses the correct reactants, but 30 kJ mol⁻¹ is below the activation energy.
B: has sufficient energy but uses HI, which is a product in the forward reaction, instead of I₂.
C: uses the correct reactants, but 44 kJ mol⁻¹ is below the activation energy.
Remember: the collision energy must meet or exceed the activation energy.

Question 3

Multiple choiceHard1 mark

In a sealed vessel, the concentrations in the reversible system N2(g)+3H2(g)⇌2NH3(g)\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 KcK_c

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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: Kc=[NH3]2[N2][H2]3K_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 NH3\mathrm{NH_3} so concentrations would change.

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

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