Question

For a gascous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction (2NO(g) +0,(8) 2NO,(8) the standard change in Gibbs free energy is AG - -72.6 kJ/mol. What is AG for this reaction at 298 K when the partial pressures are Pro -0.350 atm, Po, -0.500 atm, and Pro, -0.900 atm? 63.96 kJ/mol AG -

Answer 1

Answer:-

Given:-

partial pressure of NO gas (PNO) = 0.350 atm

partial pressure of O2 gas (PO2) = 0.500 atm

partial pressure of NO2 gas (PNO2) = 0.900 atm

temperature (T) = 298 K

standard free energy change of reaction ($\Delta$G0) = - 72.6 kJ.mol-1 = - 72600 J.mol-1

free energy change of reaction ($\Delta$G) = ?

As we know that

gas constant (R) = 8.314 J.K-1.mol-1

Also we know that

2NO(g) + O2(g) $\rightarrow$   2NO2(g)

Equilibrium constant (Kp) = P2NO2 / P2NO $\times$ PO2

Equilibrium constant (Kp) = (0.900 )2 / (0.350)2 $\times$ 0.500

Equilibrium constant (Kp) = 0.81‬ / 0.1225‬  $\times$​​​​​​​ 0.500

Equilibrium constant (Kp) = 0.81‬ / 0.06125‬

Equilibrium constant (Kp) = 13.22

According to formula

$\Delta$G = $\Delta$ G0 + 2.303RTlog(Kp)

then

free energy change of reaction ($\Delta$G) = standard free energy change of reaction ($\Delta$G0) + 2.303RTlog(Kp)

free energy change of reaction ($\Delta$G) = (- 72600 J.mol-1 ) + 2.303 $\times$ 8.314 J.K-1.mol-1 $\times$ 298 K $\times$ log(13.22)

free energy change of reaction ($\Delta$G) = (- 72600 J.mol-1 ) + 2.303 $\times$ 8.314 J.K-1.mol-1 $\times$ 298 K $\times$ 1.12

free energy change of reaction ($\Delta$G) = (- 72600 J.mol-1 ) + 6390.55 J.mol-1  

free energy change of reaction ($\Delta$G) = - 72600 J.mol-1 + 6390.55 J.mol-1​​​​​​​  

free energy change of reaction ($\Delta$G) = - 66209.45‬ J.mol-1  

free energy change of reaction ($\Delta$G) = - 66.21 kJ.mol-1  (i.e the answer)