CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The dissociation constant of NH4OH at 25oC is Kb=X×106 , if ΔHo and ΔSo for the given changes are as follows: NH4+H+NH+4;ΔHo=52.2kJmol1;ΔSo=+1.67JK1mol1 H2OH++OH;ΔHo=56.6kJmol1;ΔSo=78.2JK1mol1. Value of X is=5Y+2. What is the value of Y?

Open in App
Solution

NH4OH+H+NH+4;ΔHo=52.2kJmol1
Adding H2OH++OH;ΔHo=+56.6kJmol1
NH3+H2ONH+4+OH;ΔHo=+4.4kJmol1
Similarly, ΔSo for the change =76.53JK1 mol1
or for the change:
NH3+H2ONH+4+OH;ΔHo=4.4kJmol1
and ΔSo=76.53JK1 mol1
Now, we have ΔGo=ΔHoTΔSo
ΔGo=4.4(76.53×103)×298 or
=27.21kJ mol1
Also, ΔGo=2.303RT logKb
27.21=2.303×8.314×103×298×logKb
Kb=1.7×105
so Y=3.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Acids and Bases
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon