Given the following thermodynamics properties: Calculate delta H degree, delta G

Question

Given the following thermodynamics properties: Calculate delta H degree, delta G degree, and K_p for the formation of methanol at 25degreeC according to the following reaction. Predict whether or not the reaction is spontaneous at 25degreeC if carried out under standard conditions. CO_(g) + 2H_2(g) rightarrow CH_3OH_(g) Predict whether or not the reaction is spontaneous at 500.K if: (i) it is carried out under standard pressure conditions; (ii) it is carried out under the following pressure conditions: P_h2 = 60 atm; P_co = 30 atm, and P_ch3oh = 10 atm.

Explanation / Answer

Change in Gibbs free energy of the reaction = Gibbs free energy change of products- Gibbs free energy change of reactants

1* change in gibbs free energy of CH3OH- { 1* change in gibbs free energy of CO + 2Change in Gibbs free energy change of H2} = 1*(-201)+110.5 =-90.5 Kj/mol

Similalry entopry change = 1*240-(1*198+2*131) =-220 J/kmol

Enthalpy change = Gibbs free energy chagne- T * entropy change= -90.5*1000+298.15*220=-24907 J/mol = -24.907 Kj/mol

The reaction is spontaneous at 25 deg.c

Change in Gibbs free energy= -RTlnKp

-90.5*1000= -8.314*298.15*lnKp

lnKp=90.5*1000/8.314*298.15 =36.51

Kp =7.179*1015

At 500 K Change in gibbs free energy = -90.5*1000+ 500*220=19500 J/mol

The reaction is not spontaneous at 500K

For PH2= 60 atm and PCO= 30 atm and PCH3OH= 10 atm

lnKp = 10/(30*60*60)= 9.259*10-5

delG =-RT lnKp

delG is –ve the reaction is spontaneous

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