Calculate ΔrG(in kJ/mol) for given reaction at 300 KN2(g) + O2(g) ⇌ 2NO(g)the partial pressure of N2, O2 & NO are 10–4 bar, 10–2 bar and 10–5 bar respectively when mixed together at 300 K.Given : 2.303 RT = 5750 J/molAt 300 K standard enthalpy of formation of NO(g) = 90.5 kJ mol–1,Standard molar entropy data at 300 K :S°(N2, g) = 190 J K–1 mol–1S°(NO, g) = 210 J K–1 mol–1S°(O2,g) = 205 J K–1 mol–1 ?

To calculate ΔrG (in kJ/mol) for the given reaction at 300 K, we need to use the formula:

ΔrG = ΔrG° + RTlnQ

Where:

• ΔrG° is the standard Gibbs free energy change
• R is the universal gas constant
• T is the temperature in Kelvin
• Q is the reaction quotient

First, we need to calculate ΔrG°. This is given by the formula:

ΔrG° = ΔrH° - TΔrS°

Where:

• ΔrH° is the standard enthalpy change
• ΔrS° is the standard entropy change

Given that the standard enthalpy of formation of NO(g) is 90.5 kJ mol–1, and the standard molar entropy data at 300 K are S°(N2, g) = 190 J K–1 mol–1, S°(NO, g) = 210 J K–1 mol–1, and S°(O2,g) = 205 J K–1 mol–1, we can calculate ΔrH° and ΔrS° as follows:

ΔrH° = 2(90.5 kJ mol–1) - 0 - 0 = 181 kJ mol–1

ΔrS° = 2(210 J K–1 mol–1) - (190 J K–1 mol–1 + 205 J K–1 mol–1) = 25 J K–1 mol–1

Therefore, ΔrG° = 181 kJ mol–1 - 300 K * 25 J K–1 mol–1 = 181 kJ mol–1 - 7.5 kJ mol–1 = 173.5 kJ mol–1

Next, we need to calculate Q. This is given by the formula:

Q = (pNO)^2 / (pN2 * pO2)

Given that the partial pressures of N2, O2, and NO are 10–4 bar, 10–2 bar, and 10–5 bar respectively, we can calculate Q as follows:

Q = (10–5 bar)^2 / (10–4 bar * 10–2 bar) = 10^-10

Finally, we can calculate ΔrG using the given value of 2.303 RT = 5750 J/mol:

ΔrG = 173.5 kJ mol–1 + (5750 J/mol * ln(10^-10)) = 173.5 kJ mol–1 - 13.8 kJ mol–1 = 159.7 kJ mol–1

Therefore, ΔrG for the given reaction at 300 K is 159.7 kJ mol–1.