Question

At $\,60^\circ C\,$, dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere

Answer 1

Hint:Gibbs free energy is a thermodynamic potential that can be used at a constant temperature and pressure to measure the limit of reversible work that can be done by a thermodynamic system. The shift that happens during a reaction in the free energy of the system determines the equilibrium between the two driving forces that decide if a reaction is spontaneous.
Formula used:
The standard free energy change is nothing but Gibbs energy ($\Delta G$).
$\Delta G = - 2.303RT\log kp$
Where,
$R$ is rate law constant
\[T\] is temperature at kelvin
${k_p}$ is equilibrium constant calculated from partial pressure
Mole fraction $\,X = \dfrac{{{n_i}}}{{{n_{total}}}}\,$
Where, $\,{n_i} = \,$ number moles of solute and $\,{n_{total}} = \,$ total number of moles of the solution
Partial pressure $ = {P_i} = X \times {P_{total}}\,$
Where, $\,{P_i} = \,$ partial vapor pressure, $\,X = \,$ Mole fraction and $\,{P_{total}} = \,$ Total pressure
Equilibrium constant $\,{K_p} = \dfrac{{{{({{\text{P}}_{\text{C}}})}^c}{{({{\text{P}}_{\text{D}}})}^d}}}{{{{({{\text{P}}_{\text{A}}})}^a}{{({{\text{P}}_{\text{B}}})}^b}}}\,$ for a general reaction $\,aA(g) + bB(g) \leftrightharpoons cC(g) + dD(g)\,$ Where $\,P\,$ is the partial pressure of each component

Complete step by step answer:
Let us analyze the given data;
Given that
$\,T = 60 + 273 = 333K\,$
\[p\left( {total} \right) = 1atm\]
Now, let us see the dissociation of dinitrogen tetroxide;
${N_2}{O_4} \rightleftharpoons 2N{O_2}\,$
Here one mole of dinitrogen tetroxide is getting converted into two moles of nitrogen dioxide.
It is given that, $\,\,50\\% \,$ dinitrogen tetroxide dissociates to give $\,\,50\\% \,$ of nitrogen dioxide, This is represented as $\,\dfrac{{50}}{{100}}\,\,$
Therefore, number of moles of the compounds after this dissociation is as follows;
Number of moles of ${N_2}{O_4} = \dfrac{{50}}{{100}} = 0.5mol\,$
Number of moles of $\,2N{O_2} = 2 \times \dfrac{{50}}{{100}} = 2 \times 0.5 = 1\,mol\,$
Therefore, total number of moles $\, = 0.5 + 1 = 1.5moles\,$
Now, let us calculate the mole fractions;
$\,X = \dfrac{{{n_i}}}{{{n_{total}}}}\,$
Where, $\,{n_i} = \,$ number moles of solute and $\,{n_{total}} = \,$total number of moles of the solution

 Calculating mole fraction of ${N_2}{O_4}$
We have $\,{n_i} = 0.5,\,{n_{total}} = 1.5\,$
Substituting we get;
${X_{{N_2}{O_4}}} = \dfrac{{0.5}}{{1 + 0.5}} = 0.33\,$
Calculating mole fraction of $N{O_2}$
$\,\,{n_i} = 1,\,{n_{total}} = 1.5\,$
${X_{N{O_2}}} = \dfrac{1}{{1 + 0.5}} = 0.66\,$
Now, let us calculate the partial pressures;
Partial pressure $ = {P_i} = X \times {P_{total}}\,$
Where, $\,{P_i} = \,$ partial vapor pressure, $\,X = \,$Mole fraction and $\,{P_{total}} = \,$ Total pressure
Calculating partial pressure of ${N_2}{O_4}$
We have $\,X = 0.33,{P_{total}} = 1\,$
Substituting this we get;
$\,{P_{{N_2}{O_4}}} = 0.33 \times 1 = 0.33atm\,$
Calculating partial pressure of $N{O_2}$
We have $\,X = 0.66,{P_{total}} = 1\,$
Substituting this we get;
$\,{P_{N{O_2}}} = 0.66 \times 1 = 0.66\,atm\,$
Now let us calculate $\,{K_p}\,$
Equilibrium constant$\,{K_p} = \dfrac{{{{({{\text{P}}_{\text{C}}})}^c}{{({{\text{P}}_{\text{D}}})}^d}}}{{{{({{\text{P}}_{\text{A}}})}^a}{{({{\text{P}}_{\text{B}}})}^b}}}\,$ for a general reaction $\,aA(g) + bB(g) \leftrightharpoons cC(g) + dD(g)\,$
Here, ${N_2}{O_4} \rightleftharpoons 2N{O_2}\,$
$\,{K_p} = \dfrac{{{{({P_{N{O_2}}})}^2}}}{{{P_{{N_2}{O_4}}}}}\,$
Substituting ${P_{{N_2}{O_4}}}$and ${P_{N{O_2}}}$values in $\,{K_p}\,$
We get $\,{K_p} = \dfrac{{{{0.66}^2}}}{{0.33}} = 1.33atm\,$
Substituting whole value in
$\Delta G = - 2.303RT\log kp$
We have $\,R = 0.0821J{K^{ - 1}}mol{e^{ - 1}}\,$
$\,T = 333K\,$and $\,{K_p} = 1.33atm\,$
Substituting in the equation above, we get;
$\,\Delta G = - 2.303 \times 0.0821 \times 333 \times \log (1.33)\,$
$\Delta G = - 7.79atm.litre\,$
Hence standard free energy change is $\, - 7.79atm.litre\,$.

Additional information:
-Mole fraction is the ratio of the number of moles of one component of a solution
${K_P}$ is the equilibrium constant calculated from the partial pressures of a reaction equation
-Partial pressure that a gas in a mixture of gases would exert if it occupied the same volume as the mixture at the same temperature.
-Heat shifts the equilibrium in favor of $\,N{O_2}\,$ and the tube becomes darker. Since the formation of $\,{N_2}{O_4}\,$ is an exothermic reaction, lowering the temperature shifts the equilibrium in favor of colorless $\,{N_2}{O_4}\,$.In a dry ice - acetone bath, the $\,{N_2}{O_4}\,$ will crystallizes as a white solid.
-Nitrogen dioxide is a reddish- brown gas while $\,\,{N_2}{O_4}\,$ is colorless.
-According to the thermodynamic data for this system, the dimerization of $\,N{O_2}\,$ is an exothermic reaction.
-The equilibrium shifts when the temperature changes inside the vessel

Note: For understanding many chemical systems, knowledge of equilibrium constants is important, as in biochemical processes such as haemoglobin oxygen transport in the blood and acid-base homeostasis in the human body. The numerical value of an equilibrium constant is obtained by allowing a single reaction to proceed to equilibrium and then calculating the amounts of each product involved in that reaction. The ratio is determined between the product concentrations and the reactant concentrations. Since the concentrations are determined at equilibrium, for a given reaction, the equilibrium constant stays the same regardless of the initial concentrations.