Biochemistry Essay, Calculations and worksheet on here provided below.This helps to get in line with Biochemistry.Loads of description that can help in course work PDF

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Biochemistry Essay, Calculations and worksheet on here provided below.This helps to get in line with Biochemistry.Loads of description that can help in course work...

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SECTION 1 (AC1.1 / AC1.2) You will need to answer each of the questions in this section. You must include your full workings of your answers within your submission. 1. You have been given a sample of unknown molarity. Calculate the molarity of a solution which has been prepared by dissolving 8.75 moles of sodium chloride in enough water to produce a solution of 6.22l. M=moles of solute(mol)/litres of solution(l)= 8.75/6.22=1.41 mol/L. 2. You have a sample which consists of 428g sodium hydroxide (NaOH) dissolved in enough water to make a final solution of 6.4l. What is the molarity? M=moles of solute(mol)/litres of solution(l)= 428/6.4= 66.875 (66.87) mol/L. 3. You have been given a sample of 4.85 moles of glucose (C6H12O6) and you want to make a 0.75M solution with it. What will be the final volume of solution? M=moles of solute(mol)/litres of solution(l)=4.85/0.75=6.46666667 (6.46) mol/L. 4. 37.0cm³ of a sodium hydroxide (NaOH) solution was pipetted into a conical flask and titrated with a standard solution of 0.3 mol dm-3 hydrochloric acid according to the following equation: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) A universal indicator solution was used for the titration and it was found that 22.0 cm3 of the acid (HCl) was required to neutralise the alkali (NaOH). Calculate the molarity of the sodium hydroxide. Moles=molarity * volume (in dm3=cm3/100) HCl moles=0.3*(22.0/1000)=0.0066 HCl= 1:1 (0.0066 mol NaOH in 22.0cm3) 1dm3=1000cm3: 0.006 * (1000/25.0)=0.24 mol NaOH in 1dm3 Mr (NaOH)= Sodium=23, Oxygen=16, Hydrogen=1 23+16+1=40 Molarity=0.24 mol NaOH in 1dm3

SECTION 2 (AC2.1 / AC2.2) (A) Imagine you have been given a series of materials to react together and the following enthalpy change data has been recorded. Explain the terms exothermic and endothermic, and identify which of the following are exothermic and endothermic reactions. Ensure that you explain your reasoning. Your explanation should be a maximum of 500 words.

Exothermic reactions release energy during a reaction, whilst endothermic reactions take in energy. Enthalpy change can be measured in a lab and is shown with Delta symbol and H. Delta H being negative means reactants have lost energy resulting in an exothermic reaction. Delta H being positive means reactants have gained energy and an endothermic reaction has taken place. Regarding reactions above: 1- shows a positive enthalpy change (endothermic reaction), 2-shows a negative enthalpy change (exothermic reaction), 3-shows a negative enthalpy energy (exothermic reaction), 4shows a positive enthalpy charge (endothermic reaction). KJ per mol show the energy loss/gain in the reaction. Exothermic reactions allow surroundings to gain energy/feel warmer hence the loss of system energy. The energy decrease is shown on the right of equation product. Endothermic reactions cause a loss in energy to surroundings resulting in a colder temperature and energy increase to the system. Energy value is seen on the left of equation product. (B)You have been asked to calculate the enthalpy change in the following reaction: 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s) The reaction occurs in the following stages:

Note stages 2 and 3 look the same, but the reaction is cooling from 1700oC to 25oC and hence the enthalpy change there. Calculate the overall enthalpy change for the reaction. You must show your workings within your submission, and explain your reasoning. Your explanation should be a maximum of 300 words.

Hess’s Law says when reactants A are converted to Products B the enthalpy change will be the same no matter how many steps. The image below outlines a direct conversion and a two-step process. Due to the location of the reactants and enthalpy being the same, the enthalpy charge is the same. This Law is followed to calculate the charge, this is shown as the stage process in the question.

Clark, J. (2013). Hess's Law and enthalpy change calculations

An enthalpy change is when measure of a heat change is made at constant pressure. No matter what steps are taken the enthalpy change will be the same if initial and final conditions are the same. The formula for enthalpy calculation is: DeltaH= H products-H reactants. Delving deeper enthalpy is H, sum of internal energy is U, product of pressure P and volume V. This results in an equation of H=U+PV. In the above reaction pressure remains constant meaning and heat is released into surrounding making it exothermic. Add calculations

2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l) + Al2O3(s) + 2Fe(l) → Al2O3(s) + 2Fe(s) ________________________________ 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s) ΔHr = (-732.5 kJ mol–1) + (-27.6 kJ mol–1) + (-91 kJ mol–1) = -851 kJ mol–1 SECTION 3 (AC3.1 / AC3.2) (A)After experimentation you have determined the following rate equations:

Identify the order of reaction for each of these rate equations. You must show your workings within your submission, and explain your reasoning. Your explanation should be a maximum of 300 words. Rate reactions are rate of reaction and concentration resulting in information gain regarding the mechanism of said reaction. Factors which can change the rate of reaction include: temperature, pressure, particle size and a catalyst being present. Within a reaction a reactant and product are present, the reactant being used up or products produced is the reaction rate. Looking at the above equations K is the constant in the reaction. The small number outside the brackets illustrates the order of that reactant. Where no number is shown the concentration of said reactant does not affect

the rate of reaction. Where a 1 is illustrated the reaction rate will be doubled in line with the reactant being double. A two can be shown and will have a result of quadrupling reaction rate as it illustrates squared as appose to concentration. Using this information, the first-rate reaction shows one H2 reactant (power of 0) and one NO reactant (squared). These two numbers are first orders and once added calculate the second order reaction (0+2=2). Regarding the second equation, H2 is to the power of 2 and O2 is to the power of 0 (first order), resulting in an overall reaction of 2+0=2 (second order). Rate equation three has both reactants to the power of 0 resulting in a zero-order reaction.

Each reactant listed in the rate equation have an exponent of either 0, 1, or 2 (above 2 is very rare). That exponent denotes the order of that reactant. Looking at each exponent: - zero means that the concentration for that reactant has no bearing on the rate of reaction. - A one means that increasing the concentration of this reactant will increase the rate of the reaction in a linear way (doubling the reactant doubles the rate). - A two means that the rate of the reaction will increase by the square of the increased concentration (doubling the reactant will increase the rate by four times). - Zero order reactants are often not listed in the rate equation, since any number to the zeroth power is equal to one. The overall order of a reaction is the sum of each reactants' orders. Add the exponents of each reactant to find the overall reaction order. This number is usually less than or equal to two https://www.chemguide.co.uk/physical/basicrates/orders.html First reaction: 1st order w.r.t H2; 2nd order w.r.t. NO; 1 +2 = 3rd order overall Second reaction: 2nd order w.r.t. H2; 1st order w.r.t. O2 1 +2 = 3rd order overall Third reaction: 1st order w.r.t H2; 1st order w.r.t NO; 1 +1 = 2nd order overall (B) Through experimentation, you note that at high temperatures ethyl chloride produces HCl and ethylene by the following reaction: CH3CH2Cl(g) → HCl(g) + C2H4(g) Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride (CH3CH2Cl). You must show your workings within your submission, and explain your reasoning. Your explanation should be a maximum of 300 words.

From the workings above it can be seen that the concentration of ethyl chloride has an impact to the reaction rate: comparison of experiments 2 and 3 shows the concentration doubling the reaction rate. To work this out simply multiplying 2.4 by 2 to achieve the reaction time stated in experiment 3 has been done. To further support this experiment 1 and 4 are seen to quadruple the reaction rate (worked out by multiplying 1.4 by 4 to gain the reaction time stated in experiment 4). This information allows for it to be assumed the reaction relates to K[CH3CH2Cl] which is the rate law for a first order reaction. Therefore, this is a first order reaction which has taken place. Add calculations

CH3CH2Cl(g) → HCl(g) + C2H4(g) Rate = k[CH3CH2Cl]x [CH3CH2Cl]3 / [CH3CH2Cl]1 = 0.030 \ 0.010 = 3 R3 / R1 = 4.8 × 10-8 / 1.6 × 10-8 = 3 3x = 3 x=1 Rate = k[CH3CH2Cl] first order reaction

SECTION 4 (AC4.1 / AC4.2) (A) For the following reactions, determine the equilibrium constant equation. You must show your workings within your submission.

Equilibrium constant equations: Within time a reversible reaction reaches an equilibrium which results in reactions to the right becoming balanced by the reactions on the left. Proportions of materials on both sides remain constant. The above are all examples of this. An equilibrium constant can be identified by the use of the formula: aA+bB+cC xX+yY+zZ. The uppercase letter is in reference to the chemical and the lower case in reference to atoms/molecules present in chemical. The placing of ‘eqm’ remains in the same place when the temperature is constant. Kc is used to show the equilibrium constant. In relation to the equations, equation 1 shows no substances to powers so it is assumed they are to the power of 1 establishing the Kc reaction: [H2O][CO]/ [CO2][H2]. Equation 2, there are powers shown. There are more than one of the chemicals. Kc=[H2O]2[N2]/[NO]2[H2]2. Equation 3 includes ions and components in varying states. As silver and copper are in solid form but the ions in solution form, the solids are to be left out of the equilibrium equation. Kc=[Cu2+]/[Ag2+]. In equation 4 Kc=[Hl]2/[H2][I2]. The customers answer is fine, I don’t see any mistakes

(B) Read the following relating to the Haber process: “The raw materials for this process are hydrogen and nitrogen. Hydrogen is obtained by reacting natural gas - methane - with steam, or through the cracking of oil. Nitrogen is obtained by burning hydrogen in air. Air is 80% nitrogen; nearly all the rest is oxygen. When hydrogen is burned in air, the oxygen combines with the hydrogen, leaving nitrogen behind. Nitrogen and hydrogen will react together under these conditions: • A high temperature - about 450oC • A high pressure - about 200 atmospheres (200 times normal pressure) • An iron catalyst”

The chemical reaction which occurs during the Haber process is:

1. Explain Le Chatelier’s principle. Reversible reactions are affected by conditions such as temperature, pressure and presence of catalyst. Le Chantelier’s Principle explains that if the conditions change in a reversible reaction then the position of equilibrium moves to accommodate the change. For example, if a forward reaction is seen to be endothermic (take energy) the equilibrium will move forwards to the right so it can reduce the energy that has been put in via heat to maintain a reversible reaction.

2. Use Le Chatelier’s principle to explain what would happen if you raised the temperature of this reaction. Regarding the above reaction, based on Le Chantelier’s principle if the temperature was raised the equilibrium will shift to the left or right depending on whether the forwards reaction is endothermic or exothermic. In this reaction a bond between the two components is formed, making the forwards reaction exothermic. When additional heat is added to a forward’s exothermic reaction the equilibrium will move to the left in order to accommodate the need for a longer endothermic (reverse reaction) to take place so balance is maintained. Regarding the equilibrium constant reaction for reactants, it is a constant and does not move. Instead the top and bottom of the equations move to accommodate change and to maintain a balanced reaction. 3. Use Le Chatelier’s principle to explain what would happen if you increased the pressure of this reaction. If pressure were to increase in this reaction the equilibrium position may be altered. If pressure is increased the equilibrium will attempt to find a way to get the equation balanced again. Since the number of gas molecules has a difference of two from left to right then a pressure increase will move the equilibrium to the right which will have the result, more ammonia produced. 4. Use Le Chatelier’s principle to explain what would happen if used increased the concentration of Nitrogen at the start of the reaction. Should Nitrogen be increased at the beginning of the reaction, the equilibrium will move right to the change having been made to the left. This enables balance to be maintained. The result of this increase will be a greater amount of ammonia being produced meaning that hydrogen the quantity will decrease and more heat will be produced. Your response for this question should be a maximum of 500 words.

References:

Clark, J. (2013). Hess's Law and enthalpy change calculations. [online] Chemguide.co.uk. Available at: https://www.chemguide.co.uk/physical/energetics/sums.html [Accessed 3 Sep. 2019]. Bibliography: Chemguide.co.uk. (2019). Le Chatelier's Principle. [online] Available at: https://www.chemguide.co.uk/physical/equilibria/lechatelier.html [Accessed 6 Sep. 2019]. Clark, J. (2013). Hess's Law and enthalpy change calculations . [online] Chemguide.co.uk. Available at: https://www.chemguide.co.uk/physical/energetics/sums.html [Accessed 3 Sep. 2019]. Lister, T. and Renshaw, J. (2015). AQA Chemistry. 2nd ed. Oxford: Oxford University Press, pp.258, 298, 261, 267. Mtsu.edu. (n.d.). Equilibrium. [online] Available at: https://www.mtsu.edu/chemistry/chem1010/pdfs/Chapter%2010Equilibrium.pdf [Accessed 6 Sep. 2019]....