IB Chemistry SL R2.3 How far? The extent of chemical change questions

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Question 1

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

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: 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.

Question 2

Paper 2Hard12 marks

At 298 K, dinitrogen tetroxide gas dissociates to nitrogen dioxide gas in two sealed, rigid vessels. Each vessel initially contains only dinitrogen tetroxide. Figure 1 shows the structure of dinitrogen tetroxide.

ON+O-N+O-Odinitrogen tetroxide

The concentrations were measured over time. The final readings are treated as equilibrium concentrations for calculating KcK_\mathrm{c}

Time / sRun A [N₂O₄] / mol dm⁻³Run A [NO₂] / mol dm⁻³Run B [N₂O₄] / mol dm⁻³Run B [NO₂] / mol dm⁻³
00.08000.00000.12000.0000
200.06500.03000.09600.0480
400.05600.04800.08500.0700
600.05100.05800.08200.0760
800.05000.06000.08170.0767
1000.05000.06000.08170.0767
1200.05000.06000.08170.0767
(a)

Use Figure 1 to write a balanced equation, including state symbols, for the dissociation of dinitrogen tetroxide gas into nitrogen dioxide gas.

[2]
(b)

For Run A, calculate the ratio of the increase in [NO₂] to the decrease in [N₂O₄] between t=0t=0 and t=80 st=80\ \mathrm{s}

[1]
(c)

Write the equilibrium law expression for KcK_\mathrm{c} for the reaction in (a).

[2]
(d)

A researcher claims that KcK_\mathrm{c} is greater in Run B because its equilibrium concentration of NO₂ is greater. Determine whether the data support this claim. Include calculations for both runs and a conclusion.

[5]
(e)

A student argues that Run B cannot be at equilibrium because the concentrations of N₂O₄ and NO₂ are unequal. Evaluate this argument using the definition of dynamic equilibrium and the time data.

[2]

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