Question

Consider the reaction given below.

H₂(g) + F₂(g) 2 HF(g)

(a) Using thermodynamic data from the course website, calculate G° at 298 K.
____ kJ
(b) Calculate G at 298 K if the reaction mixture consists of 8.3 atm of H₂, 3.0 atm of F₂, and 0.26 atm of HF.
____ kJ

Answer 1

H₂(g) + F₂(g) 2 HF(g)

$\Delta$G⁰    = $\Delta$ G⁰f products - $\Delta$ G⁰f reactants

           = 2*-273.2 - (0 + 0)

           = -546.4KJ

            = -546400J

H₂(g) + F₂(g) 2 HF(g)

    Qp   = P^2HF/PH2*PF2

           = (0.26)^2/(8.3*3)

           = 0.0027

$\Delta$G    = $\Delta$ G⁰ + RTlnQ

            = -546400+8.314*298ln(0.0027)

          = -546400 + 8.314*298*-5.9145

          = -561053J

        = -561.053KJ >>>answer