B M B 211 Lecture Notes - Lecture 7: Endergonic Reaction, Enthalpy, Reaction Rate

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coralcrab720

30 Sep 2016

School

PSU

Department

Biochemistry and Molecular Biology

Course

B M B 211

Professor

Giebink Heather

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Document Summary

There are three ways to increase the rate of a chemical reaction. Enzymes (1) increasing the temperature (2) increasing the concentration of the reacting substances (3) adding a catalyst. Enthalpy (h): the heat contents of a system. Entropy (s): the measure of the system"s disorder or randomness. Gibbs free enegy (g): a measure of the free energy of a system based on h and s. When g<0 the reaction is spontaneous or exergonic (favorable, no need to manipulate) When g>0 the reaction is nonspontaneous or endergonic (not favorable) Nonspontaneous reactions occur in cells because they are coupled to favorable reactions. There is an energy barrier in chemical reactions that has to be overcome called the activation energy ( g ) Enzymes work by lowering the activation energy for a reaction and stabilizing the transition state. They increase reaction rates by bringing reactant groups into close proximity and orienting them properly for reaction.

Related Questions

Q1. Which of the following processes are spontaneous and which are non spontaneous.
a) The combustion of natural gas
b) Extraction of Aluminum metal from its ore
c) A ball rolling down a hill
d) A bike going up a hill

Q2. Predict the entropy change (positive or negative) for the following processes/reactions
a) a lake freezing
b) weeding a garden
c) N2(g) + 2 H2(g) ? N2H4 (g)
d) Mg(s) + 2H2O(l) ? Mg(OH)2 (s) + H2(g)

Q3. The element Cesium (Cs) freezes at 28.4 C and its molar enthalpy of fusion ΔH_fusion = 2.09 kJ/mol.
a) When molten Cs solidifies to Cs (s) at its normal melting point, is the entropy change ΔS positive or negative?
b) Calculate the entropy change S when 15.0 g of Cs(l) solidifies at 28.4 C.

Q4. Using the values for standard molar entropies ΔS_f° from the appendix in your textbook, calculate ΔS° for the following reactions.
a) Be(OH)₂ (s) ? BeO (s) + H2O (g)
b) Mg(OH)2 (s) + 2 HCl (g) ? MgCl2 (s) + 2H2O (l)
c) 2CH4(g) ? C2H6 (g) + H2(g)

Q5. Using the data in the appendix of your textbook, calculate the Gibbs free energy change ΔG° for the following reactions. In each case, indicate whether the reaction is spontaneous or not
a) H2 (g) + Cl2(g) ? 2HCl (g)
b) 2 POCl3 (g) ? 2 PCl3 (g) + O2 (g)

Q6. Consider the reaction 2 NO2 (g)---N2O4 (g)
a) Using the data in the appendix in your text, calculate the standard Gibbs free energy change (ΔG°) for the reaction at 298K.
b) For the above reaction, calculate the value of Keq at 298K
c) For the same reaction, calculate G at 298 K when the partial pressures for NO₂ and N2O4 are 0.40 atm and 0.36 atm, respectively.

Q7. Consider the reaction I2 (g) + Cl2 (g)----2 ICl (g) Kp = 81.9 at 298 K
a) Calculate the standard free energy change (ΔG°)
b) What will be the Free energy change when the above reaction is at equilibrium
c) Calculate G for the reaction at 298 K when PICl = 2.55 atm, PI2=0.325 atm and PCl2 = 0.221 atm

Q8. Consider the decomposition of Lead carbonate PbCO3 (s) ? PbO(s) + CO2 (g)
a) Calculate the Gibbs free energy change (ΔG°) for the reaction at298 K.
b) Calculate G for the reaction at 400 K.
c) Write down the equilibrium expression and calculate the value of Keq at 298 K
d) Calculate the partial pressure of CO2 at 298 K. (Hint: Look at the equilibrium expression and the value of Keq at 298 K)

Q9. For the reaction: 2Al(s) + 3/2 O2(g) ? Al2O3(s), ΔH^o_rxn = -1617 kJ/mol and ΔH^o_rxn = -1577 kJ/mol. Use this data to calculate ΔS^o_rxn reaction at 298 K.

Q10. Calculate the ΔG^o for the following reaction using the data from your text:
2SO2(g) + O2(g) ? 2SO3(g)

Q11. Using the thermodynamic data available from the Appendix(?Goâs), calculate ΔG^o rxn and calculate the equilibrium constant(K) for the reaction at 25 oC:
CaCO3(s) ? CaO(s) + CO2(g)

Q12. What is the minimum temperature required for the spontaneous conversion of CCl4(l) to CCl4(g) when ΔH°(vap) is 57.3 kJ/mol andΔS°(vap) is 164 J/(mol K)? reaction: CCl4(l) ? CCl4(g)

Q13. Given the following
Fe2O3(s) + 3CO(g) . ? 2Fe(s) + 3CO2(g) ΔG° = 29.4 kJ
3Fe2O3(s) + CO(g) . ? 2Fe3O4(s) + CO2(g) ΔG° = 61.6 kJ
Calculate ΔG° for Fe(s) + Fe2O3(s) + CO2(g) ? Fe3O4(s) +CO(g)

Q14. Calculate ΔG° rxn for 2CO(g) + O2(g) ? 2CO2(g) and then calculate ΔG for the reaction at 298K; if the pressure of CO =20. atm,P O2 = 10. atm and P CO2= 0.50 atm. Will increasing the pressure of the reactants make the reaction more or less spontaneous? Explain using your calculations and Le Chatelierâs principle.

Q15. Write the reaction for photosynthesis (production ofC6H12O6(s) and oxygen from carbon dioxide and water as a gas).Calculate ?H, ? S, and ?G for the reaction at standard state. Is the reaction spontaneous?

blackferret405

Photosynthesis is the process by which plants effectively capture energy from the sun and store it for later use by organisms. It's the fundamental process that enables the entire food chain. Actual photosynthesis is a highly complex process involving many steps and many different chemicals, but we can get an idea of what's happening to thermodynamic quantities (enthalpy, Gibbs free energy, and entropy) using simpler reactions.

Let's consider a toy model in which an input of energy (for example from the sun) is be used to combine water and carbon dioxide to create an organic molecule (formaldehyde) and oxygen, effectively "storing" that energy for use later. The simplest possible such process is given by:

CO₂ + H₂O â O₂ + CH₂O

The table lists the enthalpy of formation, ÎH_f and the entropy S for 1 mol of each of the substances involved in this reaction at 298 K and 10⁵ Pa.

	ÎH_f (kJ)	S (kJ/K)
CH₂O	-115.9	0.2189
H₂O	-241.82	0.1888
CO₂	-393.51	0.2138
O₂	0	0.2051

Answer the following questions assuming that 1 mol of formaldehyde molecules are formed.

1. What is the change in enthalpy of this reaction?
ÎH = ____ kJ

Does this change in enthalpy represent energy that must be added to the system, or energy that is released from the system?
-Energy must be added or Energy is released?

2. What is the change in entropy for this reaction?
ÎS = ____ kJ/K

Does the entropy increase or decrease?
-Entropy increases or Entropy decreases?

At 298 K, what is the heat transfer that corresponds to this change in entropy?
Q = ____ kJ
Does this heat transfer represent energy that is added to the system, or energy that is released from the system?
--Energy is added Energy is released ?

3. What is the change in Gibbs free energy for this reaction?
ÎG = _____ kJ

The Gibbs free energies of formation, ÎG_f, are: -228.57 kJ for water; -394.36 kJ for CO₂; and 0 kJ for O₂. What is the Gibbs free energy of formation for formaldehyde?
ÎG_f (formaldehyde) = _______ kJ

4. Using the formation of formaldehyde as a model, comment on photosynthesis and the formation of sugars using what we know about entropy and free energy.

Model 1 â Spontaneous Processes Process Description Change in Enthalpy (âH) Change in Entropy (âS) Spontaneous? A Two pure â Homogeneous substances mixture ~ 0 Increasing Yes B Solute on Solute on one side of â both sides of membrane membrane ~ 0 Increasing Yes C Polypeptide â Individual chain amino acids Exothermic (negative) Increasing (positive) Yes D C3 H8 + 5O2 â 3CO2 + 4H2 O Exothermic Increasing Yes E 6CO2 + 6H2 O â C6 H12O6 + 6O2 Endothermic Decreasing No F Glucose â Starch Endothermic Decreasing No G Water â Ice Exothermic Decreasing Below 0 Â°C, yes Above 0 Â°C, no H Cold water (25 Â°C) â Hot water (60 Â°C) Endothermic Increasing Below 60 Â°C, no Above 60 Â°C, yes

Some of the processes in Model 1 are spontaneousâthat is, they will occur without any additional work being done on the system. For example, a solute will diffuse across a membrane until the concentrations of both sides are equal. However, glucose will not spontaneously form from carbon dioxide and water in the atmosphere. The Second Law of Thermodynamics states that a process will be spontaneous when it results in an increase of total entropy in the universe. In other words, either the system or the surroundings must have an increase in entropy, or both. (Please note that the term âspontaneousâ does not imply that the change will happen quickly. Rusting is spontaneous under the right conditions, but it can still occur very slowly.)

13. Predict the change in entropy of the surroundings for an endothermic reaction.

B M B 211 Lecture Notes - Lecture 7: Endergonic Reaction, Enthalpy, Reaction Rate

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