Gibbs free energy question

Question

At 25C, the equilibrium partial pressures for the reaction

3A(g)+4B(g)yields 2C(g)+D(g)

were found to be PA=5.30 atm, PB=4.95 atm, PC=5.17 atm, and PD=4.56 atm.

What is the standard change in Gibbs free energy of this reaction at 25C?

Answer 1

Hello,

You need to use the equation: (delta G) = (standard delta G) + (R)(T)(lnQ)

where Q = [(PC^2)(PD)] / [(PA^3)(PB^4)] and delta G = 0 since the reaction is at equilibrium

Q is the reaction quotient that provides a measurement of the relative quantities of products and reactants present in a reaction mixture for a reaction at a particular point in time. This applies to problems involving partial pressures.

Therefore, (standard delta G) = -(R)(T)(lnQ)

Plug 'n chug!

(standard delta G) = -(8.314 J/K*mol) (298K) {ln[(5.17)^2(4.56)] / [(5.30)^3(4.95)^4]}

(standard delta G) = -3.378 J/mol

Make sure this is the units that your teacher wants the answer in. Reach out to me or schedule a session online if you have any more chem questions.

Best,

Christopher