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|Delta(f)H|" of "(H(2)O,l)gt|Delta(f)H|"...

`|Delta_(f)H|" of "(H_(2)O,l)gt|Delta_(f)H|" of "(H_(2)O,g)`
`DeltaH_("condensation")` is negative.

A

If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-13

B

If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-13

C

If STATEMENT-1 is TRUE and STATEMENT-2 is FALSE

D

If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE

Text Solution

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The correct Answer is:
A
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Assertion (A): The enthalpy of formation of H_(2)O(l) is greater than that of H_(2)O(g) . Reason (R ) : Enthalpy change is negative for the condensation reaction H_(2)O(g) rarr H_(2)O(l) .

C_(2)H_(6)(g)+3.5O_(2)(g)to2CO_(2)(g)+3H_(2)O(g) DeltaS_("vap")(H_(2)O,l)=x_(1)" "cal K^(-1) (boiling point =T_(1) ) DeltaH_(f)(H_(2)O,l)" "=x_(2) DeltaH_(f)(CO_(2))" "=x_(3) DeltaH_(f)(C_(2)H_(6))" "=x_(4) Hence, DeltaH for the reaction is :

The heat of formation (DeltaH_(f)^(@)) of H_(2)O(l) is equal to :

C_(2)H_(6)(g)+3.5)_(2)(g)rarr2CO_(2)(g)+3H_(2)O(g) DeltaS_(vap)(H_(2)O,l)=x_(1)calK^(-1) (boiling point =T_(1) ) DeltaH_(f)(H_(2)O,l)=x_(2) DeltaH_(f)(CO_(2))=x_(3) DeltaH_(f)(C_(2)H_(6))=x_(4) Hence, DeltaH for the reaction is

H_(2)O(l)toH_(2)O(g) , DeltaH=+43.7 kJ H_(2)O(s)toH_(2)O(l) , DeltaH=+6.05kJ The value of DeltaH_((sublimation)) of ice is