What will be the value of Δ G and Δ G ° for the reaction, A + B leftharpoons C + D at 27° C for which K=102 ?

Question

What will be the value of Δ G and Δ G ° for the reaction, A + B leftharpoons C + D at 27° C for which K=102 ? (A) Δ G = 0 ; Δ G° = - 11.48 kJ mol-1 (B) Δ G = 0 ; Δ G° = 11.48 kJ mol-1 (C) Δ G° = - 11.48 kJ mol-1; Δ G = 0 (D) Δ G° = 11.48 kJ mol-1; Δ G = 0

Tardigrade Admin · Accepted Answer

Δ G ° of a reaction is given by the expression. Δ G ° =-2.303 RT log K ...(i) Given, K =102 T =27° C or 300 K on putting the values in Eq (i) Δ G °=-2303 × 8.314 JK -1 mol -1 × 300 × log 102 Δ G °=-1148.28 J mol -1 or -11.48 kJ mol -1 Now, as the reaction is at equilibrium, hence Δ G=0.

Q. What will be the value of $Δ G$ and $Δ G^{\circ}$ for the reaction, $A + B ⇌ C + D$ at $2 7^{\circ} C$ for which $K = 1 0^{2}$ ?

$Δ G = 0; Δ G^{\circ} = - 11.48 k J m o l^{- 1}$

$Δ G = 0; Δ G^{\circ} = 11.48 k J m o l^{- 1}$

$Δ G^{\circ} = - 11.48 k J m o l^{- 1}; Δ G = 0$

$Δ G^{\circ} = 11.48 k J m o l^{- 1}; Δ G = 0$

Q. What will be the value of ΔG and ΔG∘ for the reaction, A+B⇌C+D at 27∘C for which K=102 ?

ΔG=0;ΔG∘=−11.48kJmol−1

ΔG=0;ΔG∘=11.48kJmol−1

ΔG∘=−11.48kJmol−1;ΔG=0

ΔG∘=11.48kJmol−1;ΔG=0

ΔG∘ of a reaction is given by the expression. ΔG∘=−2.303RTlogK ...(i) Given, K=102 T=27∘C or 300K on putting the values in Eq (i) ΔG∘=−2303×8.314JK−1mol−1×300×log102 ΔG∘=−1148.28Jmol−1 or −11.48kJmol−1 Now, as the reaction is at equilibrium, hence ΔG=0.

Q. What will be the value of $Δ G$ and $Δ G^{\circ}$ for the reaction, $A + B ⇌ C + D$ at $2 7^{\circ} C$ for which $K = 1 0^{2}$ ?

$Δ G = 0; Δ G^{\circ} = - 11.48 k J m o l^{- 1}$

$Δ G = 0; Δ G^{\circ} = 11.48 k J m o l^{- 1}$

$Δ G^{\circ} = - 11.48 k J m o l^{- 1}; Δ G = 0$

$Δ G^{\circ} = 11.48 k J m o l^{- 1}; Δ G = 0$