KVPY SX SB Exam 19-Nov-2017 Question Paper

Question : 107 of 120

Marks: +1, -0

The maximum work (in kJ mol

^{–1}

) that can be derived from complete combustion of 1 mol of CO at 298 K and 1 atm is
[Standard enthalpy of combustion of
CO = –283.0 kJ mol

^{–1}

; standard molar entropies at 298 K ;

S_{O_2}

= 205.1 J mol

^{–1}

_{CO}

= 197.7 J mol

^{–1}

S_{O_2}

= 213.7 J mol

^{–1}

]

Solution:

{CO}+\;{1}/{2} {O}_{2} → {CO}_{2}

Now,

∆ S^{0}=S_{C O 2}-[S_{C O}+\;{1}/{2} S_{O 2}]

∆ S^{0}=213.7-[197.7+\;{1}/{2} 205.1]

∆ S^{0}=-8 6 . 5 5 J \/ m o l \/ K

∆ H^{0}=-283.0 {kJ} \/ {mol}=-283.0 {kJ} \/ {mol} × 1000 {J} \/ {kJ}=-283,000 {J} \/ {mol}

Again,

∆ G^{0}=∆ H^{0}-T ∆ S^{0}

∆ G^{0}=-283,000 {J} \/ {mol}-[298 {K} ×-86.55 {J} \/ {mol} \/ {K}]

∆ G^{0}=-2 5 7 2 0 8 J \/ m o l

∆ G^{0}=-257208 {J} \/ {mol} × \;{1 {kJ}}/{1000 {J}}

∆ G^{0}=-2 5 7 J \/ m o l

The maximum work

=∆ G^{0}=-2 5 7 J \/ m o l

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