Question

Predict the ΔHrxn for the following reaction: Rb(g) + Na+(g) → Rb+(g) + Na(g)

	ΔHrxn < 0 because the IE1 for Rb is less than the IE1 for Na
	ΔHrxn > 0 because the IE1 for Rb is less than the IE1 for Na
	ΔHrxn < 0 because the IE1 for Na is less than the IE1 for Rb
	ΔHrxn < 0 because the IE1 for Na+ is less than the IE1 for Rb+
	ΔHrxn > 0 because the IE1 for Na+ is less than the IE1 for Rb+

Answer 1

In a group from top to bottom, IE decreases.

IE1 for Rb is less than the IE1 for Na.

Rb---------> Rb+ + e-         IE1 = 403 kJ/mol ----------- Eq (1)

Na ----------> Na+ + e-     IE1 = 496 kJ/mol ------------- Eq (2)

Given reaction,

Rb(g) + Na+(g) ? Rb+(g) + Na(g)

Take Eq(1) same as above

Rb---------> Rb+ + e-         IE1 = 403 kJ/mol        

Take reverse of Eq (2)

Na+ + e- ------------> Na   IE1 = - 496 kJ/mol

Hence,

Rb---------> Rb+ + e-         IE1 = 403 kJ/mol        

Na+ + e- ------------> Na   IE1 = - 496 kJ/mol

--------------------------------------------------------------------

Add above two equations

Rb(g) + Na+(g) ? Rb+(g) + Na(g)      

?Hrxn = 403 kJ/mol - 496 kJ/mol = -93 kJ/mol

Therefore,

Ans: ?Hrxn < 0 because the IE1 for Rb is less than the IE1 for Na