KERBOODLE PRACTICE QUESTIONS: ES Flashcards
Look at the two reactions of chlorine with ethene:
Reaction 1: C2H4 + Cl2 –> C2H4Cl2
Reaction 2: C2H4 + Cl2 –> C2H3Cl + HCl
Which of the following answers is correct about the atom economies of these reactions in terms of the organic product
A) R1 - 80%
R2 - 50%
B) R1 - 100%
R2 - 37%
C) R1 - 100%
R2 - 63%
D) R1 - 40%
R2 - 100%
C) R1 - 100%
R2 - 63%
Which is the correct answer for the electrolysis of aqueous potassium iodide?
A) Cathode: Potassium
Anode: Iodine
B) Cathode: Hydrogen
Anode: Oxygen
C) Cathode: Iodine
Anode: Potassium
D) Cathode: Hydrogen
Anode: Iodine
D) Cathode: Hydrogen
Anode: Iodine
K salts give H2 at cathode; iodides give I2 at anode
What are the correct half equations for the electrolysis of molten sodium chloride?
A) Cathode: Na+ + e- –> Na
Anode: Cl- –> Cl + e-
B) Cathode: 2H+ + 2e- –> H2
Anode: Cl- –> 1/2Cl2 + e-
C) Cathode: Na+ –> Na + e-
Anode: Cl2 –> 2Cl- + 2e-
D) Cathode: Na+ + e- –> Na
Anode: 2Cl- –> Cl2 + 2e-
D) Cathode: Na+ + e- –> Na
Anode: 2Cl- –> Cl2 + 2e-
Products are sodium and chlorine
Which answer correctly describes the halogen elements at room temperature?
A) Chlorine: green gas
Bromine: brown gas
Iodine: purple gas
B) Chlorine: colourless solution
Bromine: brown solution
Iodine: brown solution
C) Chlorine: green gas
Bromine: red liquid
Iodine: grey solid
D) Chlorine: yellow-green gas
Bromine: brown liquid
Iodine: purple solid
C) Chlorine: green gas
Bromine: red liquid
Iodine: grey solid
Sodium bromide is reacted with silver nitrate solution.
The result is…
A) a white precipitate that is soluble in dilute ammonia solution
B) a cream precipitate that is soluble in concentrated ammonia solution
C) a yellow precipitate that is insoluble in ammonia solution
D) a white precipitate that is soluble in concentrated ammonia solution
B) a cream precipitate that is soluble in concentrated ammonia solution
Which of the following will react with sodium iodide to produce the purest sample of hydrogen iodide?
A) concentrated sulfuric acid
B) dilute sulfuric acid
C) dilute hydrochloric acid
D) phosphoric acid
D) phosphoric acid
An aqueous solution of chlorine is added to an aqueous solution of sodium iodide. Some cyclohexane is added, forming the upper layer. Which of the following is the correct observation?
A) There is no reaction
B) The upper layer goes purple and the aqueous layer goes brown.
C) Both layers go brown.
D) The lower layer is brown and the upper layer is yellow.
B) The upper layer goes purple and the aqueous layer goes brown.
When sulfuric acid reacts with a bromide, which of the following are correct?
- *Statement 1:** hydrogen bromide is produced
- *Statement 2:** sulfur dioxide is produced
- *Statement 3:** bromine is produced
A) 1,2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 3 are correct
D) Only 1 is correct
A) 1,2 and 3 are correct
Which of the following are true about hydrogen chloride, hydrogen bromide and hydrogen iodide?
- *Statement 1:** they all react with ammonia
- *Statement 2:** they are all acidic
- *Statement 3:** they all reduce sulfuric acid
A) 1,2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 3 are correct
D) Only 1 is correct
B) 1 and 2 are correct
HCl does not reduce sulfuric acid (H2SO4)
A solution of copper sulfate is electrolysed with copper electrodes. Which of the following is true?
- *Statement 1:** Copper is transferred from the anode to the cathode.
- *Statement 2:** Copper is plated on the anode.
- *Statement 3:** Sulfur dioxide is produced at the anode.
A) 1,2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 3 are correct
D) Only 1 is correct
D) Only 1 is correct
Copper is plated on the cathode, no SO2 is produced at all
In the manufacture of bromine from sea water:
Step 1: Chlorine is bubbled through sea water containing a very dilute bromide solution to release bromine.
Step 2: Air is blown through to produce bromine vapour
Step 3: The vapour is mixed with sulfur dioxide and passed into water:
Br2 + SO2 + ………. –> 2HBr + ……….
Step 4: Steam and chlorine are blown through to release bromine from the hydrogen bromide
Step 5: The bromine is dried using concentrated sulfuric acid.
Write an ionic equation for the reaction that occurs in both steps 1 and 4. Which property of the halogens does this reaction illustrate?
Cl2 + 2Br- –> 2Cl- + Br2
Ignore state symbols
Chlorine/Chloride is a better oxidising agent than bromine/bromide
In the manufacture of bromine from sea water:
Step 1: Chlorine is bubbled through sea water containing a very dilute bromide solution to release bromine.
Step 2: Air is blown through to produce bromine vapour
Step 3: The vapour is mixed with sulfur dioxide and passed into water:
Br2 + SO2 + ………. –> 2HBr + ……….
Step 4: Steam and chlorine are blown through to release bromine from the hydrogen bromide
Step 5: The bromine is dried using concentrated sulfuric acid.
Suggest why is it necessary to produce bromine in Step 1 and then again in Step 4.
Because the bromine in Step 1 is too dilute.
In the manufacture of bromine from sea water:
Step 1: Chlorine is bubbled through sea water containing a very dilute bromide solution to release bromine.
Step 2: Air is blown through to produce bromine vapour
Step 3: The vapour is mixed with sulfur dioxide and passed into water:
Br2 + SO2 + ………. –> 2HBr + ……….
Step 4: Steam and chlorine are blown through to release bromine from the hydrogen bromide
Step 5: The bromine is dried using concentrated sulfuric acid.
On what property of bromine does Step 2 depend?
volatile or low boiling point
In the manufacture of bromine from sea water:
Step 1: Chlorine is bubbled through sea water containing a very dilute bromide solution to release bromine.
Step 2: Air is blown through to produce bromine vapour
Step 3: The vapour is mixed with sulfur dioxide and passed into water:
Br2 + SO2 + ………. –> 2HBr + ……….
Step 4: Steam and chlorine are blown through to release bromine from the hydrogen bromide
Step 5: The bromine is dried using concentrated sulfuric acid.
Suggest the appearance of the gas stream after Step 2.
brown or red (brown-red)
In the manufacture of bromine from sea water:
Step 1: Chlorine is bubbled through sea water containing a very dilute bromide solution to release bromine.
Step 2: Air is blown through to produce bromine vapour
Step 3: The vapour is mixed with sulfur dioxide and passed into water:
Br2 + SO2 + ………. –> 2HBr + ……….
Step 4: Steam and chlorine are blown through to release bromine from the hydrogen bromide
Step 5: The bromine is dried using concentrated sulfuric acid.
Use oxidation states to complete and balance the equation in step 3 and explain your reasoning.
Br2 + SO2 + 2H2O → 2HBr + H2SO4.
Two bromine atoms go from zero to -1.
so S must be oxidised from +4 to +6.
Chlorine is made by electrolysing an aqueous solution of sodium chloride.
Give the half equations for the reactions at the positive and negative electrodes during this electrolysis.
negative electrode (cathode): 2H<sup>+</sup> + 2e<sup>-</sup> --\> H<sub>2</sub> positive electrode (anode): 2Cl<sup>-</sup> --\> Cl<sub>2</sub> + 2e<sup>-</sup>
Chlorine reacts with cold aqueous sodium hydroxide as follows:
Cl2 + H2O ⇌ HCl + HClO
Write the oxidation states underneath the chlorine atoms in Cl2, HCl, and HClO. What is being reduced and what is being oxidised in this equation?
oxidation state of Cl in Cl2 = 0
oxidation state of Cl in HCl = -1
oxidation state of Cl in HClO = +1
Chlorine is both reduced and oxidised
Give the systematic name of HClO
chloric(I) acid
Hydrogen chloride is made industrially by the reaction:
H2(g) + Cl2(g) ⇌ 2HCl(g)
The reaction can reach dynamic equilibrium. Explain the meaning of the term dynamic equilibrium.
when the rate of the forward reaction = the rate of the backward reaction
in a closed system
concentrations of products and reactants are constant
Hydrogen chloride is made industrially by the reaction:
H2(g) + Cl2(g) ⇌ 2HCl(g)
Write the equation for the equilibrium constant Kc of the reaction
[HCl]2 /[H2] [Cl2]
Hydrogen chloride is made industrially by the reaction:
H2(g) + Cl2(g) ⇌ 2HCl(g)
Kc= 2 x 1033 for this reaction at 298K. What conclusion can be drawn about the composition of an equilibrium mixture at 298K.
nearly all product (HCl)
Hydrogen chloride is made industrially by the reaction:
H2(g) + Cl2(g) ⇌ 2HCl(g)
The reaction is exothermic. Discuss the effect on an equilibrium mixture of, separately, changing the hydrogen concentration, varying the temperature and varying the pressure.
pressure - no effect
equal moles on each side
increased temperature - moves position of equilibrium to the left (reverse for lowered temperature)
forward reaction - exothermic
more H2 - position of equilibrium moves to the right (reverse for less H2)
reference to opposing change or Le Chatelier for temp
use of Kc for more H2
increased temperature - more reactants/less product (HCl)
adding hydrogen H2 and HCl greater, Cl2 less
The chemist Max Bodenstein investigated the equilibrium shown below:
H2(g) + I2(g) ⇌ 2HI(g)
He allowed mixtures of known masses of hydrogen and iodine to react in sealed tubes at high temperatures until equilibrium had been established. Then he rapidly cooled the tubes and analysed the iodine present with sodium thiosulfate.
Suggest why the flasks are rapidly cooled.
so the position of equilibrium does not change
The chemist Max Bodenstein investigated the equilibrium shown below:
H2(g) + I2(g) ⇌ 2HI(g)
He allowed mixtures of known masses of hydrogen and iodine to react in sealed tubes at high temperatures until equilibrium had been established. Then he rapidly cooled the tubes and analysed the iodine present with sodium thiosulfate.
Describe how a sealed tube could be investigated to measure the mass of iodine it contains
Dissolve contents of tube in water
Titrate with sodium thiosulfate … of known concentration
mass I2 = (0.5 x moles thiosulfate x 253.8)g
The chemist Max Bodenstein investigated the equilibrium shown below:
H2(g) + I2(g) ⇌ 2HI(g)
He allowed mixtures of known masses of hydrogen and iodine to react in sealed tubes at high temperatures until equilibrium had been established. Then he rapidly cooled the tubes and analysed the iodine present with sodium thiosulfate.
From such an experiment at 500K , the masses of substances in a 100cm3 tube were found to be:
H2 = 0.20g
I2 = 25.38g
HI = 161.15g
Calculate a value for Kc for the reaction at 500K.
[HI] = 161.15/127.9 (= 1.26)
[H<sub>2</sub>] = 0.2/2 (= 0.1) AND [I<sub>2</sub>] = 25.38/253.8 (= 0.1)
Use of [HI]2 /[H2] [I2]
answer 159 (±2)
Give the systematic name of KIO3
potassium iodate(V)
20.0cm3 of a solution of 0.002mol dm-3 KIO3 is reacted with excess iodide ions in the presence of acid.
KIO3 + 5I- + 6H+ –> 3I2 + 3H2O
Calculate the volume of 0.50mol dm-3 Na2S2O3 that would react with the iodine formed. Give your answer to a suitable number of significant figures.
n(KIO3) = 20 × 0.002/1000
= 4 x 10–4
n (Na2S2O3) = 6 × (4 x 10–4)
vol Na2S2O3 = (6 × (4 x 10–4)) x 1000/0.5
= 4.8
to 3 sf with units 4.80cm3
20.0cm3 of a solution of 0.002mol dm-3 KIO3 is reacted with excess iodide ions in the presence of acid.
KIO3 + 5I- + 6H+ –> 3I2 + 3H2O
A teacher tells a student that this is not a very satisfactory titration result. Suggest why the teacher says this and suggest what the student could do to improve the titration result without changing the apparatus used.
titre can be measured to ±0.1
larger percentage uncertainty for smaller titre
use more conc KIO3 or more dilute thiosulfate
