Question

Solid N2O5 reacts with water to form liquid HNO3. Consider the reaction with all substance in their standard states. (a.) Is the reaction spontaneous at 25 degrees Celsius? (b.) The solid decomposed to NO2 and O2 at 25 degrees Celsius. Is the decomposition spontaneous at 25 degrees C? At what temperature is it spontaneous?

Answer 1

Part (a)

The balanced equation for the formation of liquid nitric acid from the reaction of solid N₂O₅ and water is:

$N_20_5(s) + H_20(L) \to2 HNO_3(l)$

The equation for the change in free energy of a reaction is given by:

$\Delta G_{rxn}^{0}=\sum m\Delta G_{f(products)}^{0}-\sum n\Delta G_{f(reactants)}^{0}$

Here, m and n moles of the individual species given as coefficients in the balanced equation 

$\Delta G^{0}= [2*-79.914]-[1*114+1*-237.192]=-37 kJ$

So the standard free energy change of the reaction is -37 kJ. As the sign of pGois negative the reaction is spontaneous at 298 K. 

Part (b)

The balanced equation for the decomposition of solid N₂0₅, to NO₂ and 0₂ is:

$2 N_2O_5 (s) \to NO_2(g)+O_2(g)$

The equation for the change in free energy of a reaction is given by:

$\Delta G_{rxn}^{0}=\sum m\Delta G_{f(products)}^{0}-\sum n\Delta G_{f(reactants)}^{0}$

Here, m and n moles of the individual species given as coefficients in the balanced equation 

$\Delta G^{0}= [4*51+1*0]-[2*114]=-24 kJ$

So the standard free energy change of the reaction is -24 kJ. As the sign of pGois negative the reaction is spontaneous at 298 K. 

To determine the temperature at which the reaction is spontaneous we first determine the $\Delta S^0$ . and $\Delta H^0$ values for this reaction as follows: 

The standard entropy change of the reaction is:

$\Delta S^0 _{rxn}=[(4)(239.9 )+ (1)(205.0 ]—[(2)(178] =808.6 J/K$

The standard enthalpy change of the reaction is:

$\Delta H^0_{rxn}= [(4)(33.2)+ (1)(0)] —[(2)( —43.1)] =219kJ$

$\frac{219}{808.6}*1000=270.8K$

Therefore the reaction will become spontaneous at 270.8 K. 

Part (c)

The balanced equation for the decomposition of solid N₂0₅, to NO₂ and 0₂ is:

$2 N_2O_5 (s) \to NO_2(g)+O_2(g)$

The equation for the change in free energy of a reaction is given by:

$\Delta G_{rxn}^{0}=\sum m\Delta G_{f(products)}^{0}-\sum n\Delta G_{f(reactants)}^{0}$

Here, m and n moles of the individual species given as coefficients in the balanced equation 

$\Delta G^{0}= [4*51+1*0]-[2*114]=-24 kJ$

So the standard free energy change of the reaction is -24 kJ. As the sign of pGois negative the reaction is spontaneous at 298 K. 

To determine the temperature at which the reaction is spontaneous we first determine the $\Delta S^0$ . and $\Delta H^0$ values for this reaction as follows: 

The standard entropy change of the reaction is:

$\Delta S^0 _{rxn}=[(4)(239.9 )+ (1)(205.0 ]—[(2)(346] =472.6 J/K$

The standard enthalpy change of the reaction is:

$\Delta H^0_{rxn}= [(4)(33.2)+ (1)(0)] —[(2)( 11)] =110.8kJ$

$\frac{110.8}{472.6}*1000=234K$

Therefore the reaction will become spontaneous at 234K.