For a reversible reaction at 298 K the equilibrium constant K is 200. What is the value of Δ G° at 298 K?

Question

For a reversible reaction at 298 K the equilibrium constant K is 200. What is the value of Δ G° at 298 K? (A) -13.13 kcal (B) -0.13 kcal (C) -3.158 kcal (D) -0.413 kcal

Tardigrade Admin · Accepted Answer

Applying Δ G°=-2.303 RT× logK =-2.303×2×298× log 200 =-3158.4 cal = -3.158 kcal

Q. For a reversible reaction at $298 K$ the equilibrium constant $K$ is $200$ . What is the value of $Δ G^{\circ}$ at $298 K$ ?

$- 13.13 k c a l$

$- 0.13 k c a l$

$- 3.158 k c a l$

$- 0.413 k c a l$

Applying $Δ G^{\circ} = - 2.303 RT \times l o g K$
$= - 2.303 \times 2 \times 298 \times l o g 200$
$= - 3158.4 c a l = - 3.158 k c a l$

Q. For a reversible reaction at 298K the equilibrium constant K is 200. What is the value of ΔG∘ at 298K?

−13.13kcal

−0.13kcal

−3.158kcal

−0.413kcal