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MCAT Sample Questions : Physical Sciences
Several features of sulfuric acid are given below.
Preparation of Sulfuric Acid
Sulfuric acid is commonly prepared by the combustion of elemental sulfur to sulfur dioxide, followed by the catalytic oxidation of sulfur dioxide to sulfur trioxide. Sulfur trioxide is then absorbed into a 98% aqueous solution of H2SO4, and water is added to maintain a 98% concentration. SO3 reacts with the water in the aqueous solution according to Reaction 1.
SO3(g) + H2O(l) = H2SO4(l)
Concentrated sulfuric acid is 98% H2SO4 and 2% water by mass. It has a density of 1.84 g/mL and a boiling point of 338oC.
Preparation of Other Acids
HCl(g) and HNO3(l) may be prepared by the reaction between sulfuric acid and the sodium salt of the corresponding conjugate base (Cl- or NO3-, respectively).
formation of SO2
Sulfuric acid forms SO2 gas when it reacts with several compounds. For example, I2 and SO2 are formed when I- reacts with concentrated H2SO4; Br2 and SO2 are formed when Br- reacts with concentrated H2SO4. Cu+ and SO2 are formed in hot solutions of Cu(s) in H2SO4. This last reaction is unusual, because most metals react with solutions of H2SO4 to form hydrogen gas and a metal sulfate.
Following are some sample questions on this passage:
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MCAT Sample Question Number :
- When sulfuric acid reacts with copper, how does the oxidation number of the sulfur change?
- From +4 to +6
- From +6 to +4
- From +6 to +8
- From +8 to +6
Explanation: The passage states that sulfuric acid reacts with Cu(s) to produce Cu+ and SO2. Thus, sulfuric acid is converted into sulfur dioxide, or H2SO4= SO2. The oxidation number of sulfur in H2SO4 can be found by assigning oxidation numbers of +1 for hydrogen and -2 for oxygen. For the formula H2SO4 to be neutral, the sum of the oxidation numbers must be zero. If x is the oxidation number of sulfur in H2SO4, then: 2(1) + 4(- 2) + x = 0, and x = +6. Likewise, for SO2: 2(- 2) + x = 0, and x = +4. The change in oxidation number is from +6 to +4, which is answer choice B.
- The apparatus shown below can be used to prepare HNO3(boiling point = 86oC)
The yield of HNO3 collected in the tube can be maximized by maintaining the temperatures of the flask and tube, respectively, at:
- 0oC and 100oC.
- 100oC and 0oC.
- 350oC and 150oC.
- 350oC and 100oC
Explanation: The boiling point of HNO3 is given in the question as 86oC. Because HNO3 must boil out of the flask and be trapped in the tube, the temperature of the flask must be above the boiling point of HNO3 (i.e., < 86oC) and the temperature of the tube must be less than the boiling point of HNO3 (i.e, < 86oC). Answer B meets these criteria; the other answers do not. Thus, answer choice B is the best answer.
- Which of the following is the balanced equation describing the combustion of elemental sulfur?
- 2 H2S + 3 O2 = 2 SO2 + 2 H2O
- H2S + 2 O2 = SO3 + H2O
- 2 SO3 = 2 S + 3 O2
- S + O2 = SO2
Explanation: The combustion of elemental sulfur involves a reaction between oxygen (O2) and sulfur (S). [Note: Though sulfur exists as S8 molecules, its reactions are normally written in terms of its empirical formula S.] Only Answer D shows such a reaction. Thus, answer choice D is the best answer.
- In the second step of preparing H2SO4from elemental sulfur (the catalytic oxidation of SO2), which strategy is most likely to increase the yield of SO3 formed?
- Reducing the reaction temperature
- Reducing the reaction pressure
- Removing SO3 from the reaction mixture
- Removing O2 from the reaction mixture
Explanation: The reaction involves the formation of gaseous SO3 from gaseous O2 and gaseous SO2.
O2(g) + 2 SO2(g) = 2 SO3(g)
According to Le Châtelier's principle, any action that causes the reaction to shift toward the right will cause O2 and SO2 to react and increase the yield of SO3. Of the four possible actions (answers A-D), the removal of SO3 as it forms will shift the reaction toward the right and is the most likely action to increase the yield of SO3. Thus, answer C is the best answer.
- If H2(g) is formed from the reaction of Fe(s) with dilute H2SO4(aq), which species acts as the reducing agent?
Explanation: The problem hypothesizes the liberation of hydrogen in accord with the unbalanced equation below.
Fe(s) + H2SO4 = H2(g)
In the conversion, hydrogen goes from an oxidation state of +1 in sulfuric acid to 0 (zero) in H2. Thus, the hydrogen in H2SO4 is reduced (i.e., it undergoes an algebraic decrease in oxidation state). Fe(s) is the reducing agent (i.e., it causes the reduction). Thus, answer choice A is the best answer.
- Which of the following species has the smallest concentration in 98% H2SO4?
Explanation: The first ionization of sulfuric acid, H2SO4, is normally 100% in water. However, under conditions of low water content, all of the H2SO4 cannot ionize. Qualitatively, the only source of SO42- is HSO4-. The Ka for the second ionization step of a parent acid is a few orders of magnitude smaller than that of the first step; therefore, SO42- must be the least abundant species, because it is only formed in the second ionization step. Thus, answer choice A is the best answer.
Quantitatively, the mass relationship in 100 g of 98% H2SO4 is 98 g H2SO4 and 2 g H2O. The molar mass of H2SO4 is 98 g/mol and of water is 18 g/mol. Thus, in 100 g of 98% H2SO4 there is one mole of H2SO4 and 2/18 or 1/9 mole of H2O. In excess water, H2SO4 would ionize completely. However, in this case (i.e., very low water content), only about 1/9 of a mole can react stoichiometrically with water to form H3O+ and HSO4-. Of this 1/9 mol, only a small fraction of the HSO4- further ionizes to H3O+ and SO42-, because HSO42- is a weak acid. Therefore, SO42- is the chemical with the lowest concentration. Thus, answer choice A is the best answer.