Question

Determine if it is Spontaneous or non-spontaneous and determine its direction.

Calculate the ∆H° and ∆S° for the following reaction.

NH4NO3(s)+H2O(I)—> NH4+(aq)+N0-3(aq)

Answer 1

$\Delta H (NH_4NO_3)=-365.56$

$\Delta H (NO_3)^-=-207.36$

$\Delta H (NH_4^+)=-132.51$

$\Delta H=1x \Delta H(NH_4^+)+1x \Delta H (NO_3^-)- \Delta H(NH_4NO_3)=-132.51-207.36-(-365.56)=25.69 kJ$

Endothermic process

$\Delta S=\Delta S(NH_4^+)+\Delta S(NO_3^--)\Delta S(NH_4NO_3)=1x113.39+1x146.44-1x151.08=108.75 J/K$

Increasing of entropy.

$\Delta G=\Delta G(NH_4^+)+ \Delta G(NO_3^-)-\Delta G(NH_4NO_3)=1x(-79.37)+1x(-111.34)-1x(-184.01)=-6.7 kJ$

Process is spontaneous.