Calculate W,ΔH and ΔU for vaporisationof 2.5 moles of water for the given reaction at 373 K.
H₂O_(l) →H₂O_(g) ΔH= 40.68 kJ

Question

Calculate W,ΔH and ΔU for vaporisationof 2 5 moles of
water for the given reaction at 373 K
H2O(l)
→H2O(g) ΔH= 40 68
kJ

Vartika Jain · Accepted Answer

Dear User,
H2O (l) → H2O(g)Here, ∆n=2 5-0=2
5                               (only gaseous moles are counted)R = 8
314JK-1mol-1T=373 KVapourisation is an expansion process, i
e
 work is done by the system here,So, w= -P∆V=-∆nRT          = -(2
5)(8
314×10-3)(373)          = -7
752 kJThe negative sign here denotes that work is done by the system
Now, we know that∆H=∆U + ∆nRTor, ∆U=∆H-∆nRTor, ∆U=(40
68-7 752)kJ = 32 928 kJ

Calculate W,ΔH and ΔU for vaporisationof 2.5 moles of water for the given reaction at 373 K. H2O(l) →H2O(g) ΔH= 40.68 kJ

Calculate W,ΔH and ΔU for vaporisationof 2.5 moles of water for the given reaction at 373 K.
H₂O_(l) →H₂O_(g) ΔH= 40.68 kJ