The equilibrium constant Kc value for equilibrium at 227°C is 1 × 1010 . If the standard enthalpy change for this is -97.95 kJ mol-1 , the standard entropy change in J K-1 mol-1 for this equilibrium at the same temperature is (R=8.3 J K-1 mol-1 and 2.303 ≃ 2.3)

Question

The equilibrium constant Kc value for equilibrium at 227°C is 1 × 1010 . If the standard enthalpy change for this is -97.95 kJ mol-1 , the standard entropy change in J K-1 mol-1 for this equilibrium at the same temperature is (R=8.3 J K-1 mol-1 and 2.303 ≃ 2.3) (A)  -5  (B)  +5  (C)  +10  (D)  -10

Tardigrade Admin · Accepted Answer

Δ G°=-2.303RT log K =-2.3×8.3×500 log(1×1010)=-95450 J Now, Δ G°=Δ H°-TΔ S ⇒ -95450=-97950-500 Δ S ⇒ Δ S=-5 J K-1 mol-1

$- 5$

$+ 5$

$+ 10$

$- 10$

$Δ G^{\circ} = - 2.303 RT l o g K$
$= - 2.3 \times 8.3 \times 500 l o g (1 \times 1 0^{10}) = - 95450 J$
Now, $Δ G^{\circ} = Δ H^{\circ} - T Δ S$
$\Rightarrow - 95450 = - 97950 - 500 Δ S$
$\Rightarrow Δ S = - 5 J K^{- 1} m o l^{- 1}$

Q. The equilibrium constant Kc​ value for equilibrium at 227∘C is 1×1010 . If the standard enthalpy change for this is −97.95kJmol−1 , the standard entropy change in JK−1mol−1 for this equilibrium at the same temperature is (R=8.3JK−1mol−1 and 2.303≃ 2.3)

−5

+5

+10

−10

ΔG∘=−2.303RTlogK =−2.3×8.3×500log(1×1010)=−95450J Now, ΔG∘=ΔH∘−TΔS ⇒−95450=−97950−500ΔS ⇒ΔS=−5JK−1mol−1

$- 5$

$+ 5$

$+ 10$

$- 10$

$Δ G^{\circ} = - 2.303 RT l o g K$
$= - 2.3 \times 8.3 \times 500 l o g (1 \times 1 0^{10}) = - 95450 J$
Now, $Δ G^{\circ} = Δ H^{\circ} - T Δ S$
$\Rightarrow - 95450 = - 97950 - 500 Δ S$
$\Rightarrow Δ S = - 5 J K^{- 1} m o l^{- 1}$