In anaerobic cells, glucose, C 6 H 12 O 6 , is converted to lactic acid in the r

Question

In anaerobic cells, glucose, C6H12O6, is converted to lactic acid in the reaction

Glucose --> 2Lactic Acid

The DHfo of glucose and lactic acid are -1273.1 and -694.0 kJ/mol, respectively. The Cp,mo for glucose and lactic acid are 219.2 and 127.6 J/mol K, respectively.

Calculate the molar enthalpy associated with the formation of lactic acids from glucose at 298 K.

What would the quantity in part (a) be if the reaction proceeded at a physiological temperature of 310 K?

Based on your answers in parts (a) and (b), how sensitive is the enthalpy of this reaction to moderate temperature changes?

Explanation / Answer

According to the question

As we know that

a)

Delta Hr0 = Delta Hf0(products) - Delta Hf0 ( reactant)

= 2 ×DeltaHf0(lactic acid)- Delta Hf0 ( Glucose)

= (2mole× (-694 KJ/mole)) -(1× -1273.1KJ/mole)

Then,

= -1388 KJ/mole + 1273.1 KJ

= -114.9KJ

Therefore,

Molar enthalphy change associated with reaction at 298K is -114 9KJ

Then,

b)

Delta H is state function we can apply Hess law

Molar heat capacity of glucose = 219.2J/mole °C

Molar heat capacity of lacticic acid = 127.6 J/mole °C

Then,

Heat released to cool glucose from 310K to 298K=1mol× (-12 K) × 219.2J/mole K = -2.630KJ

Hr at 298K = -114.9KJ

Delta Hr0 + Delta H(Cooling)= -114.9KJ + (-2.630 KJ)

=-117.53KJ

Heat required to raise the temperature of lactic acid from 298K to 310 K = 2mole ×12K × (127.6 KJ / mole K)

=3.062 KJ

Net Delta H = Delta Hr0 +Delta H(cooling of glucose)+Delta H(heating of lactic)

=-114.9KJ +(-2.630KJ) + (3.062 KJ)

=- 114.5 KJ

Therefore,

Molar enthalphy change associated with the reaction at 310K is =-114.5 KJ

Then,

c)

Comparing to values of a and b the reaction is not sensitive at moderate temperature changes.

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