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Anonymous
Anonymous asked in Science & MathematicsChemistry · 7 days ago

HEEEELLLLPPPPP!!!!!!?

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species.

For the reaction

C2H6(g)+H2(g)↽−−⇀2CH4(g)

 

the standard change in Gibbs free energy is Δ𝐺°=−69.0 kJ/mol . What is ΔG for this reaction at 298 K when the partial pressures are 𝑃C2H6=0.350 atm , 𝑃H2=0.250 atm , and 𝑃CH4=0.800 atm ?

1 Answer

Relevance
  • hcbiochem
    Lv 7
    7 days ago

    Δ𝐺 = Δ𝐺° + RT ln Q

    where Q = PCH4^2 / PC2H6 PH2

    Δ𝐺 = -69.0 kJ/mol + 8.314 J/molK (298K) ln (0.800)^2 / (0.350)(0.250)

    Δ𝐺 = -64.1 kJ/mol

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