1. Consider the oxidation of methanol, CH3OH(g) +3/2(O2)(g) = CO2(g) + 2H2O(l) A

ID: 898437 • Letter: 1

Question

1. Consider the oxidation of methanol,

CH3OH(g) +3/2(O2)(g) = CO2(g) + 2H2O(l)

At T 25 degrees Celcius

Delta H formation (CO2,g)= -393.51 kj/mol

Delta H formation ( H2O,l)= -285.85 kj/mol

Delta H formation ( CH3,OH,g)= -200.66 kj/mol

Standard Entropy S(CO2, g)= 213.74 j/K*mol

Standard Entropy S(H20, l)= 69.91 J/K *mol

Standard Entropy S(CH3OH, g)= 239.81 J/k*mol

Standard Entropy S(O2, g)= 205.14 J/K*mol

a) Using the provided information calculate delta H formation at 25 degree celcius

b) Using the provided information calculate delta S reactant at 25 degree celcius

c) Calculate the total change in entropy (system and surroundings.)

d) Calculate Delta G formation at 25 degree celcius

e) Evaluate the equilibrium constant at 25 degree celcius

a) dH = Hproduct - Hreactant

dH = (-393.51 + -285.85*2) - (-200.66 + 0) = -764.55 kJ/mol

b) dS = Sproduct - Sreact

dS = (213.74 + 2*69.91)-(239.81+3/2*205.14)= -193.96 J/Kmol

Suniverse = Ssurrounding + Ssystem

S system = dS and Ssurrounding = Q/T

Since Q = dH then Ssurr = dH/T = -764.55/(298) = -2.5656 kJ/mol = 2565.6 J/molK

Suniverse = -193.96 + 2565.6 = 2371.6 J/molK

dG = dH - TdS

dG = -764.55 - 298*-193.96/1000 = -706.75 kJ/mol

e) Find K

dG = -RT*lnK

-706.75 = -8.314*298*lnK

706.75/(8.314*298) = lnK

K = exp(0.2852) = 1.33

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