Question

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction N 2 ( g ) + 3 H 2 ( g ) − ⇀ ↽ − 2 NH 3 ( g ) the standard change in Gibbs free energy is Δ G ° = − 69.0 kJ/mol . What is ΔG for this reaction at 298 K when the partial pressures are P N 2 = 0.250 atm , P H 2 = 0.150 atm , and P NH 3 = 0.750 atm ?

Answer 1

Answer:

ΔG = -52.9 kJ/mol

Explanation:

Step 1: Data given

Temperature = 298 K

All species have a partial pressure of 1 atm

Δ G ° = − 69.0 kJ/mol

Step 2: The balanced equation

N2(g) + 3H2(g) ⇆ 2NH3 (g)

Step 3: Calculate Q

we will use the expression: ΔG = ΔG° + RT*ln(Q)

⇒with Q = the reaction coordinate: Q = (PNH3)²/ ((PN2)*(Ph2)³) = 666.67

Step 4: Calculate ΔG

So, ΔG = -69.0 kJ/mol + (0.008314 kJ/mol*K)*(298 K)*ln(666.67) = -52.9 kJ/mol

(R = the gas constant = 8.314 J/mol* K OR 0.008314 kJ/mol*K)