Question

Consider the reaction 3CH4(g)→C3H8(g)+2H2(g).
Calculate ΔG at 298 K if the reaction mixture consists of 41 atm of CH4 , 0.013 atm of C3H8 , and 2.3×10−2 atm of H2 .
Given:
ΔH for...
CH4 = -74.8 C3H8 = -103.85 H2 = 0
ΔS for...
CH4 = 0.1863 C3H8 = 0.2699 H2 = 0.13058
ΔG for...
CH4 = -50.8 C3H8 = -23.4 H2 = 0
ΔG at 298 K = 128.9 kJ

Answer 1

From the calculations performed, the free energy change for the reaction is 72 kJ/mol.

What is the equilibrium constant?

The equilibrium constant is a value that shows the extent to which reactants have been converted to products.

Given that the equation of the reaction is;

3CH4(g)→C3H8(g)+2H2(g)

Then;

PC3H8 = 0.013 atm

PH2 =  2.3×10−2 atm

PCH4 = 41 atm

Now;

ΔG = ΔG° + RTlnQ

ΔG°reaction = ΔG°products - ΔG°reactants

ΔG°reaction = [( -23.4) +2(0)] - 3(-50.8)

ΔG°reaction = 129 kJ/mol

Q = PC3H8 * PH2^2/PCH4^3

Q = 0.013 * (2.3×10−2)^2/( 41)^3

Q = 6.877 * 10^-6/68921

Q= 9.9* 10^-11

Hence;

ΔG = 129 * 10^3  + [8.314 * 298 * (ln  9.9* 10^-11 )]

ΔG = 129 * 10^3 - 57073

ΔG = 72 kJ/mol

Learn more about free energy change: https://brainly.com/question/14143095