Consider the following reaction approaching equilibrium at 27° C and 1 atm pressure A + B xrightleftharpoons[k r=102]kf=1 03C + D The standard Gibb's energy change (Δr G°) at 27° C is (-) kJ mol -1 (Nearest integer). (Given: R =8.3 J K -1 mol -1 and ln 10=2.3 )

Question

Consider the following reaction approaching equilibrium at 27° C and 1 atm pressure A + B xrightleftharpoons[k r=102]kf=1 03C + D The standard Gibb's energy change (Δr G°) at 27° C is (-) kJ mol -1 (Nearest integer). (Given: R =8.3 J K -1 mol -1 and ln 10=2.3 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

∵ Δ G o =- RT ln K eq text and K eq =( K f / K b ) ∴ K eq =(103/102)=10 ∴ Δ G =- RT ln 10 ⇒-(8.3 × 300 × 2.3)=-5.7 kJ mole -1 ≈ 6 kJ text mole -1( text nearest integer ) text Ans =6

Q. Consider the following reaction approaching equilibrium at 27∘C and 1atm pressure A+Bkf​=103kr​=102​C+D The standard Gibb's energy change (Δr​G∘) at 27∘C is (−)___ kJmol−1 (Nearest integer). (Given: R=8.3JK−1mol−1 and ln10=2.3 )

∵ΔGo=−RTlnKeq​ and Keq​=Kb​Kf​​ ∴Keq​=102103​=10 ∴ΔG=−RTln10 ⇒−(8.3×300×2.3)=−5.7kJmole−1≈6kJ mole −1( nearest integer ) Ans =6

$∵ Δ G^{o} = - RT ln K_{e q}$
$and K_{e q} = \frac{K _{f}}{K _{b}}$
$∴ K_{e q} = \frac{1 0 ^{3}}{1 0 ^{2}} = 10$
$∴ Δ G = - RT ln 10$
$\Rightarrow - (8.3 \times 300 \times 2.3) = - 5.7 k J m o l e^{- 1} \approx 6 k J$
$mole^{- 1} (nearest integer)$
$Ans = 6$