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Grade 12 Chemistry entrance exam, 2013 E.C. (2021)

Real questions from the Grade 12 university entrance exam in Chemistry, as students sat it in 2013 E.C. Every question comes with the correct answer and a worked explanation.

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  • Natural science stream

    Who sits it

  • 2013 E.C. (2021)

    Exam year

  • 69 questions with answers

    In this set

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Answer each question yourself before opening the answer. The explanations point to the exact place in the textbook, so you know which page to reread when you miss one.

Question 1

From the magnitude of the equilibrium constant, Keq one can deduce that

  1. A.large K eq value indicates that the reaction reaches equilibrium with very little reactants remaining.
  2. B.Small K eq values indicate the reaction yields very higher amounts of products near equilibrium.
  3. C.K eq will be larger when we have higher amounts of reactants over products.
  4. D.Intermediate K eq value indicates that the reaction does proceed in the forward direction.
Show the answer and explanation

Answer: A large K eq value indicates that the reaction reaches equilibrium with very little reactants remaining.

Why

A large equilibrium constant (Keq > 10^3) indicates that the forward reaction is highly favored, meaning that at equilibrium, the mixture consists almost entirely of products with very little reactants remaining.

Question 2

The statement 'If a system at equilibrium is subjected to a stress, the system will readjust itself to reduce the effects of the stress' is known as

  1. A.Avogadro's principle.
  2. B.Faradays law.
  3. C.Le-Chatelier's principle.
  4. D.Boyle's law.
Show the answer and explanation

Answer: C Le-Chatelier's principle.

Why

Le-Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to re-establish an equilibrium.

Question 3

The law of mass action states that

  1. A.The K eq expression is the ratio of the concentrations of reactants to the products raised to their coefficients.
  2. B.The rate of a reaction is directly proportional to the product of the concentration of reactants raised to the power of their respective coefficients in the balanced equation.
  3. C.For a reaction at equilibrium, the reaction quotient, Q is always less than the equilibrium constant, Keq.
  4. D.A chemical equilibrium is attained only when the reaction is started with reactants.
Show the answer and explanation

Answer: B The rate of a reaction is directly proportional to the product of the concentration of reactants raised to the power of their respective coefficients in the balanced equation.

Why

The law of mass action states that the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.

Question 4

For the reaction between gaseous NO and O 2 to for NO 2 (g), what will be the equilibrium constant expression, 2NO(g) + O 2 (g)↔ 2NO 2 (g)

  1. A.K p =(P NO2 )/(P NO )(P O2 )
  2. B.K c =(NO 2 ) 2 / (NO) 2 (O 2)
  3. C.K p =(P NO2 ) 2 (P O2 ) / (P NO2 )
  4. D.K p =(P NO2 ) 2 /(NO) 2 (O 2 )
Show the answer and explanation

Answer: B K c =(NO 2 ) 2 / (NO) 2 (O 2)

Why

The equilibrium constant expression Kc is the ratio of the molar concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. Thus, for 2NO(g) + O2(g) ⇌ 2NO2(g), Kc = [NO2]^2 / ([NO]^2 * [O2]).

Question 5

The equilibrium constant K p for the reaction 2SO 2 (g) ⇌ 2SO 3 (g) + O 2 (g) is 3.8 × 10 -3 K p at 227 °C. What is the value of K c for the reaction at the same temperature?

  1. A.4.6 × 10 -4 mol m -3
  2. B.1.8 × 10 -3 mol m -3
  3. C.2.3 × 10 -4 mol m -3
  4. D.9.1 × 10 -4 mol m -3
Show the answer and explanation

Answer: D 9.1 × 10 -4 mol m -3

Why

For the reaction 2SO3(g) ⇌ 2SO2(g) + O2(g), the change in moles of gas is Δn = 1. Using Kc = Kp / (RT) with T = 500 K and R = 0.08206 L atm/(mol K) yields approximately 9.1 * 10^-5 mol/L (often misprinted with different exponents or units as 9.1 * 10^-4 mol m^-3).

Question 6

In the Galvanic cell given in the diagram below Identify the component parts correctly labelled for the letters a, b, c, d and choose the correct one among the following a, b, c, and d , respectively

  1. A.Copper anode, Zinc cathode, voltmeter , salt bridge
  2. B.Zinc cathode, Copper cathode, Zinc anode, Copper cathode
  3. C.Copper anode, Zinc anode, CuSO 4 (sol), voltmeter
  4. D.Zinc anode, Copper cathode, ZnSO 4 (sol), salt bridge
Show the answer and explanation

Answer: D Zinc anode, Copper cathode, ZnSO 4 (sol), salt bridge

Why

In a standard Daniell (Galvanic) cell, zinc acts as the anode, copper as the cathode, zinc sulfate and copper sulfate as the electrolyte solutions, and a salt bridge connects the two half-cells.

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Chemistry papers from other years

2018 E.C. (2026)2017 E.C. (2025)2016 E.C. (2024)2015 E.C. (2023)2014 E.C. (2022)2012 E.C. (2020)2011 E.C. (2019)2010 E.C. (2018)2009 E.C. (2017)2008 E.C. (2016)2007 E.C. (2015)2006 E.C. (2014)2005 E.C. (2013)

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Are these the real 2013 E.C. Chemistry entrance exam questions?

Yes. They come from the national university entrance exam in Chemistry that students sat in 2013 E.C. (2021). Our set holds 69 of its questions, each with its answer and an explanation that points back to the textbook.

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