Question

The standard free energy of formation of NO(g) is 86.6
kJ/mol at 298 K. What is the standard free energy of
formation of NO2(g) at 298 K? (K = 1.6 10^12)
​

Answer 1

Answer:

156.22 kJmol^{-1}

Explanation:

The equation for the reaction of NO and $O_{2}$ is as follows:

$NO(g) + \frac{1}{2} O_{2} \right arrow NO_{2} (g)$

The expression to calculate the standard Gibbs free energy change for the reaction is as follows:

$\Delta G_{rxn}^{o}=-RTlnK_{eq}$

Where,

R is the gas constant.

T is the temperature.

$K_{eq}$ is the equilibrium constant.

Substitute the value in the above expression

$\Delta G_{rxn}^{o}=-(8.314 J K^{-1}mol^{-1})(298K)ln(1.6*10^{12})=-69622.312 Jmol^{-1}$

The expression to calculate the standard free energy change of the formation is as follows:

$[tex]\bigtriangleup _{f}G^{\circ}_{rxn}=\sum a\bigtriangleup G^{\circ}_{f}(products)-\sum b\bigtriangleup G^{\circ}_{f}(reactants)$

Where,

$\bigtriangleup _{f}G^{\circ}_{rxn}$ is the standard Gibbs free energy change for the reaction.

$\bigtriangleup _{f}G^{\circ}(Product)$ is the change in the standard formation constant of the reactants.

$\bigtriangleup _{f}G^{\circ}(reactants)$ is the change in the standard formation constant of the products.

a is the reactant stoichiometric coefficient.

b is the product stoichiometric coefficient.

Substitute the value in the above expression.

$[tex]\bigtriangleup _{f}G^{\circ}_{rxn}=(1)G^{\circ}_{f}(NO_{2})-[(\frac{1}{2})G^{\circ}_{f}(O_{2})+(1)G^{\circ}_{f}(NO)]$

Rearrange above equation for $NO_{2}$.

$G^{\circ}_{f}(NO_{2})=(-69622.312 Jmol^{-1})(\frac{1 kJ}{1000 J}) + [\frac{1}{2}(0) + (1)86.6 kJmol^{-1})]=156.22 kJmol^{-1}$