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Calculate Delta(r)G^(@) for (NH(4)Cl,s) ...

Calculate `Delta_(r)G^(@)` for `(NH_(4)Cl,s)` at 310K.
Given :`Delta_(r)H^(@)(NH_(4)Cl,s)` =-314 kj/mol,`Delta_(r)C_(p)=0`
`S_(N_(2)(g))^(@)=192 JK^(-1mol^(-1)),S_(H_(2)(g))^(@)=130.5JK^(-1)mol^(-1),`
`S_(Cl_(2)(g))^(@)=233JK mol^(-1), S_(NH_(4)Cl(s))^(@)=99.5JK^(-1)mol^(-1)`
All given data at 300K

A

`-198.56 kJ//mol`

B

`-426.7 kJ//mol`

C

`202.3 kJ//mol`

D

`-84.5 kJ//mol`

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta_(f)S^(@)(NH_(4)Cl, S)` at 300 K
`= S_(NH_(4)Cl(s))^(@) - [(1)/(2)S_(N_(2))^(@) + 2S_(H_(2))^(2) + (1)/(2)S_(Cl_(2))^(@)]`
`= 374 JK^(-1) mol^(-1) " " because Delta_(r)C_(p) = 0`
`because Delta_(f)S_(310)^(@) = Delta_(f)S_(300)^(@) " " = -374 JK^(-1) mol^(-1)`
`Delta_(f)H_(310)^(@) = Delta_(f)H_(300)^(@) = -314.5`
`Delta_(f)G_(310)^(@) = Delta_(f)H^(@) - 310 Delta S^(@)`
`= -314.5 - (310(-374))/(1000) = -198.56 kJ//mol`
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Calcualate Delta_(f)G^(@) "for" (NH_(4)Cl,s) at 310 K. Given : Delta_(f)H^(@)(NH_(4)Cl,s)=-314.5 KJ//"mol," "Delta_(r)C_(p)=0 S_(N_(2)(g))^(@)=192 JK^(-1), " "S_(H_(2)(g))^(@)=130.5 JK^(-1)mol^(-1), S_(Cl_(2))^(@)(g) =233 JK^(-1)"mol"^(-1)," "S^(@)NH_(4)Cl(s)=99.5 JK_(1)"mol"^(-1) All given data are at 300K.

Calculate the magnitude of standard free energy of formation of ammonium chloride at 25^(@)C (approximate integer in Kcal mol^(-1) ), the equation showing the formation of NH_(4)CI from its elements is 1/2NH_(4)(g) + 2H_(2)(g) + 1/2CI_(2)(g) rarr NH_(4)CI(s) For NH_(4)CI, DeltaH_(f)^(@) is -313 KJ mol^(01) , Also given that " "S_(N_(2))^(@) = 191.5 JK^(-1) mol^(-1)" "S_(N_(2))^(@) = 130.6 JK^(-1) mol^(-1) " "S_(CI_(2))^(@) = 223.0 JK^(-1) mol^(-1) " "S_(NH_(4)CI)^(@) = 94.6 JK^(-1)mol^(-1)

The approx value of K for a reaction AhArrB is : Given Delta_(r)H_(298K)^(@)=-54.07KJmol^(-1) , Delta_(r)S_(298K)^(@)=10Jk^(-1)mol^(-1) and R=8.314JK^(-1)mol^(-1) .

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