A-level Chemistry PapSol Q 4-7


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4 The compound chlorine dioxide, ClO2, can be prepared by the reaction shown. 
NaClO2 + 1 2Cl 2 ClO2 + NaCl 
(a) Using oxidation numbers, explain why this reaction is a redox reaction. .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [2] 
(b) The central atom in the molecule of ClO2 is chlorine. Draw the ‘dot-and-cross’ diagram for ClO2. Show outer electrons only. [2]

 
(c) The reaction between ClO2 and F2 is shown.
 
                                                          2ClO2 + F2 2ClO2F 

 The rate of the reaction was measured at various concentrations of the two reactants and the following results were obtained. 

experiment            [ClO2]/moldm–3              [F2]/moldm–3           initial rate /moldm–3 s–1 
       1                          0.010                               0.060                                  2.20 × 10–3 
       2                          0.025                               0.060                              to be calculated 
       3                    to be calculated                     0.040                                  7.04 × 10–3 

 The rate equation is rate = k [ClO2][F2]. 
(i) What is meant by the term order of reaction with respect to a particular reagent? ............................................................................................................................................. ....................................................................................................................................... [1] 

(ii) Use the results of experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k. rate constant, k = ............................. units ............................. [2] 

(iii) Use the data in the table to calculate 
● the initial rate in experiment 2, initial rate = ............................. moldm–3 s–1 



● [ClO2] in experiment 3. [ClO2] = ............................. moldm–3 [2] 



(d) (i) What is meant by the term rate-determining step? ............................................................................................................................................. ....................................................................................................................................... [1] 
(ii) The equation for the reaction between ClO2 and F2 is shown. 

                                                         2ClO2 + F2 2ClO2F 

                                                            rate = k[ClO2][F2] 
The mechanism for this reaction has two steps. 
Suggest equations for the two steps of this mechanism, stating which of the two steps is the rate-determining step. 
step 1 .................................................................................................................................. 
step 2 .................................................................................................................................. 
rate-determining step = ............................. [2] 

(e) By considering the rate equation, explain why the rate increases with increasing temperature. .................................................................................................................................................... .............................................................................................................................................. [1] 

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5 (b) State and explain how the solubility of the Group 2 hydroxides varies down the group. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [4] 
(c) Magnesium hydroxide is sparingly soluble in water. The concentration of its saturated solution at 298K is 1.7 × 10–4moldm–3. 
(i) Write an expression for the solubility product, Ksp, of Mg(OH)2. 


Ksp =  





(ii) Calculate the value of Ksp for Mg(OH)2 at 298K and state its units. 








Ksp = ............................. units ............................. [2]  

(d) The temperature at which the Group 2 hydroxides and carbonates start to decompose increases down the group. Suggest an explanation for this trend in the decomposition temperature of the Group 2 hydroxides. .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [2]  

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6(b) Magnesium(I) chloride, MgCl, is an unstable compound and readily decomposes as shown.

                                               2MgCl(s) Mg(s) + MgCl 2(s) 

Use the following data to calculate the enthalpy change of this reaction. 

 ∆H˚fMgCl(s) = –106kJmol–1 
 ∆H˚fMgCl 2(s) = –642kJmol–1 




enthalpy change = ............................. kJmol–1 

(c) (i) The equation for which ΔH is the lattice energy for MgCl is shown. 

                                                               Mg+(g) + Cl – (g) MgCl(s) 
Use the equation, the following data, and relevant data from the Data Booklet to calculate a value for the lattice energy of MgCl. You might find it helpful to construct an energy cycle. 

electron affinity of Cl(g) = –349kJmol–1 
enthalpy change of atomisation of Mg(s) = +147kJmol–1 
enthalpy change of formation of MgCl(s) = –106kJmol–1 
SOLUTION:

    


ΔH atm(Mg) + ΔH atm(Cl) + IE(Mg) + EA(Cl) + LE(MgCl) = ∆H˚f(MgCl)
Put values and solve for unknown.




lattice energy MgCl = ............................. kJmol–1 [3] 

(ii) Suggest how the lattice energies of MgCl 2 and NaCl will compare to that of MgCl. Explain your answers. 
MgCl 2 and MgCl
SOLUTION:

Mgcl2 has higher lattice energy than MgCl because Mg2+ has higher charge than Mg+. Greater the attraction more difficult it is to move the ions apart. So, more energy is required and we call this energy lattice energy.

NaCl and MgCl
SOLUTION:

NaCl has higher lattice energy because Na+ has a smaller ionic radius than Mg+. So, the charge density is greater for Na+ as amount of charge is the same but area over which the charge is distributed is smaller for Na+.


 
(d) Define the term electron affinity. .................................................................................................................................................... .............................................................................................................................................. [2] 

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7 An electrochemical cell consists of a half-cell containing V3+(aq) and V2+(aq) ions and another half-cell containing VO2 +(aq) and VO2+(aq) ions. 

(a) (i) Use data from the Data Booklet to calculate a value for the Ecell 

There is an easy way to find this:
Ecell = (E of the equilibrium shifted to the right) - (E of the equilibrium shifted to the left)

The equilibrium with E more positive than the other is shifted to the right in cell and vice versa.

So,
Ecell = 1-(-0.26)



 ............................. V [1] 


(ii) Write the ionic equation for the cell reaction. 
Add the two equilibrium equations and than cancel the same terms on the both sides of the equation to simplify. 

V3+ + e- ⇌ V2+  (Going to the left)
VO+ + 2H+ + e- ⇌ VO2+ +H2O (Going to the right)
 
V2+ +CO2+ + 2H+ ⇌ V3+ + VO2+ +H2O







[1] 
(iii) Draw a fully labelled diagram of the apparatus you could use to measure the potential of this cell. Include the necessary chemicals. [4] 


(b) Use data from the Data Booklet to predict whether a reaction might take place when the following pairs of aqueous solutions are mixed. If a reaction occurs, write an equation for it and calculate the. 

 ● V2+(aq) and Sn4+(aq) 

V3+ + 2e- ⇌ V2+    (Going to left)    E = -0.26
Sn4+ + 2e- ⇌ Sn2+(Going to right)    E = +0.15

Ecell = (E of the equilibrium shifted to the right) - (E of the equilibrium shifted to the left)
         = 0.15-(-0.26)
Equation is found by adding equilibria = 2V2+ Sn4+ ⇌ 2V3+ + Sn2+

Remember: When adding equilibrium equations, you must first know the direction in which the equilibria are going in a electrochemical cell. Then you must rearrange the equations so taht the direction of the reaction is to the right. Only after that you can add the reactants with reactants and products with products.


Does a reaction occur? ......................... equation ..................................................................................................................................... ........................................................ 

 ● VO2+(aq) and Fe3+(aq) 

1--- VO2+ + 2H+ + e– ⇌ V3+ + H2O                  E = +0.34
2--- VO2 + + 2H+ + e– ⇌ VO2+ + H2O              E = +1.00
3--- VO3 – + 4H+ + e– ⇌ VO2+ + 2H2O             E = +1.00
4--- Fe3+ + e– Fe2+                                                 E = +0.77 
5--- Fe3+ + 3e– Fe                                                   E = –0.04

Now we have a couple of equations but we have to pick only two. So, lets give reasons to eliminate the rest of the three we don't need.
From 4 and 5 equation 4 is has more positive E so the reaction is more likely to happen. So, we drop 5.
 We reject  equation 1 too as if we add the equations 4 and 1 we don't get VO2+ and Fe3+ as the reactants. 
Please add these equations by yourself before reading ahead
We can take any of the remaining equations involving VO2+

Now for the second part :
Find the Ecell of the reaction and we get:
Ecell = 0.77-(1)=-0.23

As it is negative the reaction not likely to happen.



Does a reaction occur? ......................... equation ..................................................................................................................................... ........................................................ [3] 



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