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Special Note: Endothermic Reactions. As long as the total entropy is slightly positive, endothermic reactions will go forward Spontaneity is determined by the increase in the entropy of the system, NOT a decrease in the ENERGY (heat) of the system!. . Equilibrium.

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special note endothermic reactions
Special Note: Endothermic Reactions
  • As long as the total entropy is slightly positive, endothermic reactions will go forward
  • Spontaneity is determined by the increase in the entropy of the system, NOT a decrease in the ENERGY (heat) of the system!

equilibrium
Equilibrium
  • A system at equilibrium has no net change in products or reactants
  • This does not mean that nothing is happening!!!!
    • The Forward rate equals the Reverse rate
  • STot for systems at equilibrium equals zero

STot = 0 (at equilibrium)

gibbs free energy
Gibbs Free Energy
  • The Gibbs Free Energy, G, is a value that allows us to predict spontaneity by taking both Enthalpy AND Entropy into account

1) We know that:

STot = S + SSurr(but SSurr=- H/T)

STot = S -H/T (at constant T and P)

2) We’ll define the Gibbs free energy as:

G = H - TS

gibbs free energy1
Gibbs Free Energy

But we want to know the change in free energy when a reaction occurs, so we change it to:G = H -TS (at constant T)

But if pressure is constant, we know that:

STotSH/T

Which we can combine and rearrange into:

G = -TSTot (at constant T and P)

gibbs free energy2
Gibbs Free Energy

G = H - TS (at constant T)

G = -TS (at constant T and P)

  • A negative value of G indicates that a reaction will spontaneously occur
  • Large negative H values (like we’d have in a combustion reaction) would probably give you a large negative G
  • If TS is large and positive, the value of G may be large and negative
free energy and temperature
Free Energy and Temperature
  • Free energy decreases (becomes more negative) as temperature increases
  • At low T, Gm for solid phase is lower than that of liquid or vapour, so the solid phase is prevalent
  • As we increase T to Tfus and higher, the liquid state has a lower Gm, so it is the phase that prevails
  • As we increase T further to Tb, the gas phase has the lowest value of Gm
7 13 gibbs free energy or reaction
7.13: Gibbs Free Energy or Reaction
  • To determine the spontaneity of a reaction, we use the change in the Gibbs Free Energy, G, or the Gibbs Free Energy of Reaction

We’ve seen something like this before somewhere…

standard gibbs free energy of formation g f
Standard Gibbs Free Energy of Formation, Gf°
  • Gf° = The standard Gibbs Free Energy of reaction per mole for the formation of a compound from its elements in their most stable form.
  • Most stable form?
    • Hydrogen = ?
    • Oxygen = ?
    • Iodine = ?
    • Sodium = ?
  • Gf°is for the formation of 1 mole of product
    • Different amounts of reactants may be used…Be vigilant!

g f what does it mean
Gf°: What Does it Mean?
  • Compounds with Gf° > 0 are Thermodynamically Unstable
  • Compounds with Gf° < 0 are Thermodynamically Stable
gibbs free energy and nonexpansion work
Gibbs Free Energy and Nonexpansion Work
  • we = ‘Extra work’
    • Nonexpansion work is any kind of work other than that done against an opposing pressure
  • Stretching a spring, moving a rope, importing a sugar molecule into a cell are all examples of nonexpansion work
  • All cellular processes are examples of nonexpansion work
  • How are the Gibbs Free Energy and we related?
gibbs free energy and w e
Gibbs Free Energy and we
  • G = we
  • If we know the change in free energy, we know how much nonexpansion work can be done
  • What does this mean?
    • Let’s look at the combustion of glucose.
g and the combustion of glucose
G° and the Combustion of Glucose

C6H12O6(s) + 6 O2(g) --> 6 CO2 (g) + 6 H2O (l)

  • The G° of the reaction is -2879 kJ

For 1 mole of glucose, we get 2879 kJ of energy

Or

For 180 g of glucose, we get 2879 kJ of energy

  • To make one mole of peptide bonds, 17 kJ of work must be done.
    • If we get 2879 kJ of energy from one mole of glucose, we should be able to make 170 moles of peptide bonds

One molecule of glucose will provide enough energy to add 170 amino acids to a growing protein (in actuality, you can only add 10 amino acids)

the effect of temperature on g
The Effect of Temperature on G°
  • Remember that H° or S° is the sum of the individual enthalpies or entropies of the products minus those of the reactants
  • If we change the temperature, both are affected to the same extent, so the H° and S° values don’t significantly change
  • This is not the case with G°. Why?

G° = H° - TS°