Question

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)
ΔH∘f (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
C∘P,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

Answer 1

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

dS = S_products - S_reactants

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 = H_proudcts - H_reactants

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,

Q_surrounding = - HRxn

S_surroundings = Qsurr/T

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

dS_surroundings = -9.4050 J/mol k

D)

The entropy change of the Universe at T = 298K

Entropy of the universe:

dS_suniverse = dS_surroundings + +dS_system

   = -260.6 J/mol k - 9.4050 J/mol k

= -270.005 J/molK