Question

In: Chemistry

Consider the formation of glucose from carbon dioxide and water (i.e., the reaction of the photosynthetic...

Consider the formation of glucose from carbon dioxide and water (i.e., the reaction of the photosynthetic process): 6CO2(g)+6H2O(l)→C6H12O6(s)+6O2(g).
The following table of information will be useful in working this problem:

T= 298 K CO2(g) H2O(l) C6H12O6(s) O2(g)
ΔHf (kJ mol−1) -393.5 -285.8 -1273.1 0.0
S∘ (J mol−1 K−1) 213.8 70.0 209.2 205.2
CP,m (J mol−1 K−1) 37.1 75.3 219.2 29.4

A. Calculate the entropy change for this chemical system at T = 298 K.

B. Calculate the enthalpy change for this chemical system at T = 298 K.

C. Calculate the entropy change of the surroundings at T = 298 K..

D. Calculate the entropy change of the universe at T = 298 K.

E. Calculate the entropy change for this chemical system at 320. K .

F. Calculate the enthalpy change for this chemical system at 320. K .

G. Calculate the entropy change of the surroundings at 320. K .

H. Calculate the entropy change of the universe at 320. K .

Thank you so much

Solutions

Expert Solution

A) The entropy change for this chemical system at T = 298 K:

dS = Sproducts - Sreactants

6CO2(g) + 6H2O(l)→ C6H12O6(s) + 6O2(g).

dS = (C6H12O6 + 6*O2) - (6*CO2 + 6*H2O)

dS = (209.2 + 6*205.5) - (6*213.8 + 6*70)

dS = -260.6 J/ mol k

B)

The enthalpy change for this chemical system at T = 298 K.

Enthalpy change:

dH = Hproudcts - Hreactants

dH = (C6H12O6 + 6*O2) - (6*CO2 + 6*H2O)

dH = (-1273.1 + 6 * 0 ) - (6 * -393.5 + 6 * -285.8)

dH = 2802.7 J/mol k

C)

The entropy change of the surroundings at T = 298 K.

Entropy change in surroundings:

The temperature of surroundings does not change, so assume T = isothermal, 298 K

So,

Qsurrounding = - HRxn

Ssurroundings = Qsurr/T

dSsurrounginds = -HRxn / T = -2802.7 / (298)

dSsurroundings = -9.4050 J/mol k

D)

The entropy change of the Universe at T = 298K

Entropy of the universe:

dSsuniverse = dSsurroundings + +dSsystem

   = -260.6 J/mol k - 9.4050 J/mol k

= -270.005 J/molK


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