In anaerobic cells, glucose, C6H12O6, is converted to lactic acid in the reactio
ID: 488879 • Letter: I
Question
In anaerobic cells, glucose, C6H12O6, is converted to lactic acid in the reaction C_6H_12O_6 rightarrow 2 CH_3CH(OH)COOH. The Delta H degree_f of glucose and lactic acid are -1273. I and -694.0 kJ mol-1, respectively. The C degree_p, m for glucose and lactic acid are 219.2 and 127.6 J mol-1 K-1, respectively a. Calculate the molar enthalpy associated with the formation of lactic acid from glucose at 298 K. b. What would the quantity in part (a) be if the reaction proceeded at a physiological temperatures of 310 K? c. Based on your answers in parts (a) and (b), how sensitive is the enthalpy of this reaction to moderate temperature changes?Explanation / Answer
a) Hr0 = Hf0(products) - Hf0 ( reactant)
= 2 ×Hf0(lactic acid)- Hf0 ( Glucose)
= (2mole× (-694 KJ/mole)) -(1× -1273.1KJ/mole)
= -1388 KJ/mole + 1273.1 KJ
= - 114.9KJ
Therefore, Molar enthalphy change associated with reaction at 298K is -114 9KJ
b) Since 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
Heat released to cool glucose from 310K to 298K=1mol× (-12 K) × 219.2J/mole K = -2.630KJ
Hr at 298K = -114.9KJ
Hr0 + 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
NetH = Hr0 +H(cooling of glucose)+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
c) comparing to values of a and b the reaction is not sensitive at moderate temperature changes.
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