Core Principles Of Chemistry Assignment Sample

Introduction- Core Principles Of Chemistry

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LO1

1.

1. a) Exothermic reactions are those which would generate heat in the reaction media and on the other hand endothermic reactions are those which absorb heat and lead to progress cooling effect (Chaim and Estournès, 2018).
2. b) Activation energy is the minimum energy required to initiate a chemical reaction and formation of products from reactants.
3. c) Enthalpy diagram for an endothermic reaction is given below.

2.

1. a) Temperature and pressure are the main parameters deviating enthalpy of reaction. The standard condition for the enthalpy change is concerning 100 kPa of pressure and 298 kelvin of temperature.

b)

1. i) Standard enthalpy change of formation can be defined as the change of potential energies between products and reactants in case of formation of a certain compound. For example, if A and B react to form C then the standard enthalpy change of formation of compound C would be the difference between enthalpy of C and summation enthalpy of A and B.
2. ii) Combustion means burning of something in the presence of air then standard enthalpy change of combustion would be equal to the difference between the enthalpies of reactants and products formed. For example, combustion of A with excess oxygen would produce compound B then the change of standard combustion enthalpy would be equal to the difference of enthalpies between B and A and oxygen.

iii) Standard enthalpy change of reaction can be defined as the change of enthalpies between products and reactants.

1. c) Equation of enthalpy changes for different compounds.
2. i) Standard enthalpy changes of formation of propanol or C3H7OH -255.16 kJ/mol.

C2H4 (g) + CO/ 2H2 (g) → C3H7OH (g)

1. ii) Standard enthalpy change for combustion of propanol would be followed by the equation below.

C3H7OH (g) + 9/2 O2 (g) → 3CO2 (g) + 4H2O (g)

Standard enthalpy of combustion of 1 mole of propanol is - 2017.7 kJ under 1 atmospheric pressure and 298.15 K temperature.

iii) Standard enthalpy change for the formation of water has been described below.

H2 (g) + ½ O2 (g) → H2O (g)

The value of enthalpy change in this case would be equal to the -241.8 kJ/mol.

1. iv) Reaction for the combustion of octane has been given below.

C3H18 (g) + 25/2O2 (g) → 8CO2 (g) + 9H2O (g)

The change of enthalpy for the combustion of octane is -5074.1 kJ/mol.

3. a) Combustion of propane would follow the reaction C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)

(1) C(s) + O2(g) → CO2(g) [?H = -393 kJ/mol]

(2) H2(g) + 1/2 O2(g) → H2O(l) [?H = -286 kJ/mol]

(3) 3C(s) + 4H2(g) → C3H8(g) [ΔH = -104 kJ/mo] and reversing this would become C3H8(g) → 3C(s) + 4H2(g) and ΔH would become positive (-104 kJ/mol).

Equation 1×3 + eq 2×4 + eq 3 would be

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g) and the calculated ?H would become (-393×3 + -286×4 + 104) or -2219 kJ per mole.

1. b) Equation for the formation of ammonia has been given below

½ N2 (g) + 3/2 H2 (g) → NH3 (g)

Standard formation enthalpy of ammonia is -46 kJ/mol.

4.

1. a) 0.5 gram of hexane (C6H14) equal to 0.5/(6×12 + 14)= 0.5/86= 0.0058 moles.

Change of enthalpy can be measured for per moles of substance burnt then it would be equal to q/no of moles reacted.

Hence q= (c × m × ?T)

Therefore q= (4.18 × 200 × 28)= 23408 J = 23.4 kJ.

Ultimately, the change of combustion enthalpy for hexane would be equal to -23.4/0.0058= -4035 kJ/mol (value of enthalpy change is negative as the reaction is exothermic).

1. b) Enthalpy change from the calculation of bond energies may avoid some cases including rotational and vibrational energy of the compound and also the bond energies for different compounds are also varied.

5.

1. a) Hess’s law: Overall change of enthalpy would be equal to the enthalpy changes for each step of the reaction.
2. b) Formation of butane would follow the equation 4C(s) + 5H2 (g) = C4H10 (g)

Data equations for enthalpy of combustion:

C(s) + O2(g) = CO2 (g); ?H= -394 kJ [eq 1]

H2(g) + ½O2(g) = H2O(l); ?H= -286 kJ [eq 2]

2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 (g); ?H= - 2877 kJ [eq 3]

Multiplying equation 1 by 4, equation 2 by 5, and reversing the equation 3 we can get the proper equation of formation which has been described earlier.

Adding them would help to find out the change of enthalpy for the formation of butane and the corresponding value would become (4× -393 + 5 × -286 + 2877)= -125 kJ/mol.

1. c) NH3 (g)+ HCl (g) → NH4Cl (s)

NH3 (g) → NH3 (aq); ?H= - 8.4 kCal. [1]

HCl (g) → HCl (aq); ?H= -17.3 kCal [2]

NH3 (aq)+HCl (aq) → NH4Cl(aq); ΔH= -12.5 kCal [3]

NH4Cl (s) → NH4Cl(aq); ΔH=+3.9 kCal and reversing this would become NH4Cl (aq) → NH4Cl(s); ΔH= -3.9 kCal

Adding those equations would help to calculate the formation enthalpy of NH4Cl from NH3 and HCl and the value would become -42.1 kCal /mole or -176.1 kJ/mole.

LO2

1.

1. a) Rate of reaction means the time taken to complete any chemical reaction. It can be defined as the minimum time taken to complete any chemical reaction.
2. b) Four main factors that can affect the rate of reaction and those are reactant concentration and physical state, surface area, temperature and catalytic presence of any chemical compounds (Su et al. 2021).

2.

1. a) Both curves regarding the effect of temperature on the reaction rate have been depicted below.

Figure 2: Compare reaction rate with change of temperature

(Source: Hand drawn)

1. b) Using the Maxwell Boltzman distribution curve of reaction potential would help to describe the dependence of temperature on the rate of reaction (Wang et al. 2021). With increasing temperature of the reaction media, reaction rate increases followed by the decrease of activation energy.
2. c) The area of the graph represents the number of gas molecules present in the reaction media.
3. d) From the Maxwell Boltzman distribution curve, presence of catalysts would help to determine the rate of reaction. In the presence of any catalyst, the rate of reaction increases because it reduces the activation energy of the reactants.

 LO3

1.

1. a) Whenever a system has been described to be present at a dynamic equilibrium, the reactants and product ratio according to their strength remain in equilibrium.
2. b) Dynamic equilibrium exists when the concentration of reactants and products remain constant without escaping anything from the reaction media.
3. c) Le Chatelier principle describes the effect of temperature, pressure, concentration, and volume for a reaction system present in equilibrium (Kim et al. 2019).
4. H2O (g) + CO (g) ? H2 (g) + CO2 (g); ?H= -42.1 kJ/mol.
5. i) With increasing the concentration of the CO, the reaction would shift backward leading to formation of fewer products.
6. ii) With decreasing pressure, the reaction would tend to become more exothermic as more heat must be produced to make up the loss of heat. Therefore, the rate of reaction increases with adding more temperature in the system.

iii) If pressure increased then the reaction would shift towards the less concentrated side or the side contains less moles of substances. In this case, the number of moles of reactants and products are the same and therefore, it would not be affected by the pressure.

Industrial production of ammonia or the Haber process is the example for the compromised pressure and temperature.

N2 (g) + H2 (g) ? NH3 (g); ?H= -92 kJ/mol.

A high temperature and pressure would assist to enhance the rate of the reaction but it is costly as well. Compromise temperature of 450 degree Celsius and pressure of 200 atm are used to get higher yield at a lower cost comparatively.

LO4

1. a)
2. i) As per the Bronsted and Lowry theory an acid is a proton donor or the substance which donates protons in the aqueous solution.
3. ii) A base is a proton acceptor or the chemical compound which accepts protons from the solution.
4. b) An alkali is that chemical compound which produces hydroxide ion (OH-) in its aqueous solution.
5. c)
6. i) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

In the above reaction, HCl is the acid and NaOH is the base.

1. ii) NH3(aq) + H2O(l) → NH4 + (aq) + OH- (aq)

In the above reaction, NH3 is the base and H2O acts as an acid because NH3 accepts the proton which is donated by the water molecules.

1. d) pH scale can be used to determine any compound acid or base itself. It is scaled from 1 to 14 where 7 is the neutralize point. If the pH of the solution remains less than 7 then it is acidic and more than 7 then it would be the base itself (Agarwal, 2019).
2. e) Strong acid produces protons faster and consequent concentration of hydronium ions would increase as well. Whereas, the strong base would accept protons faster and the concentration of the hydronium ion would reduce then or the concentration of hydroxide ion increases.
3. i) HNO3 (aq) + KOH (aq) = KNO3 (aq) + H2O (aq)
4. ii) 2HCl (aq)+ CaCO3 (aq) = CaCl2 (aq) + CO2 (g) ↑+ H2O (aq)

iii) H2SO4 + 2NaOH (aq) = Na2SO4 (aq) + 2H2O (aq)

1. iv) 2HCl (aq) + Mg(s) = MgCl2 + H2 (g)↑
2. v) HCl + NH3 = NH4Cl (s)

Whenever a glass rod dipped in concrete HCl becomes white whenever it has been kept in the mouth of an ammonia gas jar.

1. vi) CH3COOH (aq) + NaOH (aq)= CH3COONa + H2O (aq)

vii) Nitric acid reacts with metal compounds differently with difference in concentration and temperature. Whenever concentrated nitric acid is added in the Zn metal, it would produce nitrogen dioxide and with decreasing concentration of the nitric acid it would generate nitric oxide, nitrous oxide, and nitrogen gas.

Zn + 4HNO3 (high conc.) = Zn(NO3)2 + 2NO2 + 2H2O

3Zn + 8HNO3 (moderate conc.) = 3Zn(NO3)2 + 2NO + 4H2O

4Zn + 10HNO3 (low conc.) = 4Zn(NO3)2 + N2O + 5H2O

LO5

1. In terms of electron transfer, oxidation and reduction reactions can be determined. The oxidation reaction is related with the electron release whereas the reduction reaction is concerned with the acceptance of electrons (Cao et al. 2021).
2. a) oxidation states of some compounds are given below.
3. i) HCl= oxidation state of Cl is -1.
4. ii) H2S= oxidation state of -2

iii) CH4= oxidation state of carbon is -4

1. iv) MgBr2= oxidation state of Mg is +2 and that of bromine is -1.
2. v) NaClO3= oxidation state of Cl is +5.
3. vi) K2SO4= oxidation state of S is +6.
4. b) Oxidation states of some ions are given below
5. i) OH-; oxidation state of oxygen in hydroxide ion= +1
6. ii) CO3 2-; oxidation state of carbon in carbonate ion= +4

iii) ClO4-; oxidation state of Cl in chlorate is = +7

1. iv) PO4 3-; oxidation state of phosphorus in phosphate ion= +5

3.

1. a) The term redox would describe the reduction and oxidation reactions present in a chemical reaction.
2. b)
3. i) Mg → Mg2+ + 2e-

The forward reaction is the oxidation reaction as magnesium releases to electrons to form a magnesium ion.

1. ii) Zn + Cl2 → ZnCl2

In this reaction Zn oxidized into ZnCl2 and on the other hand, chlorine reduces to the same product (Yuan et al. 2019). Zn is reductant as it releases electrons and chlorine is an oxidant as it takes up electrons released by Zn.

iii) Ag+ + Br- → AgBr

It is a redox reaction as reduction and oxidation occurs simultaneously. Ag takes up an electron which is released by bromide ion. Therefore, here silver ion is oxidant and bromide ion is reductant.

1. iv) 2H+ + 2e- → H2

It is a reduction reaction where two hydrogen ions take up two electrons to form a hydrogen molecule.

4.

1. a) 2Fe + 3Cl2 = 2FeCl3

Fe and Cl2 are present at zero oxidation state and after reaction the oxidation state of Fe becomes +3 and that of Cl becomes -1. That means two Fe releases a total of 6 electrons whereas the 3 chlorine molecule takes up those electrons simultaneously. Therefore, in this reaction, iron is the reductant and chlorine is the oxidant.

2Fe - 6e- → 2Fe3+

6Cl + 6e- → 6Cl-

1. b)

Zn - 2e- → Zn2+ (oxidation reaction)

Cu2+ + 2e- → Cu (reduction reaction)

Here both oxidation and reduction occur, therefore it can be said that the reaction is a redox reaction. Here, Zn is the reductant and Cu2+ is the oxidant.

References

Journals

Chaim, R. and Estournès, C., 2018. On thermal runaway and local endothermic/exothermic reactions during flash sintering of ceramic nanoparticles. Journal of materials science, 53(9), pp.6378-6389.

Torres-García, E., Ramírez-Verduzco, L.F. and Aburto, J., 2020. Pyrolytic degradation of peanut shell: activation energy dependence on the conversion. Waste Management, 106, pp.203-212.

Dey, D. and Chutia, B., 2022. Modelling and Analysis of Dusty Fluid Flow Past a Vertical Surface with Exothermic and Endothermic Kind of Chemical Reactions. In Advances in Thermofluids and Renewable Energy (pp. 45-57). Springer, Singapore.

Mentado-Morales, J., Mendoza-Pérez, G., Los Santos-Acosta, D., Eduardo, Á., Peralta-Reyes, E. and Regalado-Méndez, A., 2018. Energies of combustion and enthalpies of formation of carbon nanotubes. Journal of Thermal Analysis and Calorimetry, 131(3), pp.2763-2768.

Dar, A.A., Ahmed, A. and Reshi, J.A., 2018. Characterization and estimation of weighted Maxwell-Boltzmann distribution. Applied Mathematics and Information Sciences, 12(1), pp.193-202.

Su, X., Fischer, A. and Cichos, F., 2021. Towards Measuring the Maxwell–Boltzmann Distribution of a Single Heated Particle. Frontiers in Physics, 9, p.342.

Zmitrovich, I.V., Arefyev, S.P., Bondartseva, M.A. and Wasser, S.P., 2020. Professor Shu-Ting Chang, Cancer Mycotherapy and Le Chatelier Principle. International Journal of Medicinal Mushrooms, 22(9).

Wang, H., Wang, L., Lin, D., Feng, X., Niu, Y., Zhang, B. and Xiao, F.S., 2021. Strong metal–support interactions on gold nanoparticle catalysts achieved through Le Chatelier’s principle. Nature Catalysis, 4(5), pp.418-424.

Kim, S., Choi, H. and Paik, S.H., 2019. Using a Systems Thinking Approach and a Scratch Computer Program To Improve Students’ Understanding of the Brønsted–Lowry Acid–Base Model. Journal of Chemical Education, 96(12), pp.2926-2936.

Agarwal, P., 2019. Bronsted Lowry: Conjugate Acid Base Pair.

Liliasari, S., Nursa’adah, E. and Amsad, L.N., 2018. Describing Pre-service Chemistry Teachers’ Misconceptions of Proton Transfer in Acids-Bases Brønsted-Lowry.

Yuan, H., Peng, H.J., Huang, J.Q. and Zhang, Q., 2019. Sulfur redox reactions at working interfaces in lithium–sulfur batteries: a perspective. Advanced Materials Interfaces, 6(4), p.1802046.

Cao, X., Li, H., Qiao, Y., Jia, M., Li, X., Cabana, J. and Zhou, H., 2021. Stabilizing Anionic Redox Chemistry in a Mn?Based Layered Oxide Cathode Constructed by Li?Deficient Pristine State. Advanced Materials, 33(2), p.2004280.