For A B at 300 K, Delta G deg = +4.00 kJ mol^-1. Use R = 8.314 J mol^-1K^-1. Select the correct statement(s).

For Q = 0.100, Delta G approx -1.7 kJ mol^-1 and the net reaction goes forward

K approx 0.20, so [B]/[A] approx 0.20 at equilibrium

For Q = 0.100, Delta G approx -1.7 kJ mol^-1 and the net reaction goes forward

K approx 0.20, so [B]/[A] approx 0.20 at equilibrium

For Q = 1.00 the net reaction goes forward because B and A are at equal concentrations

At equilibrium Delta G = Delta G deg = +4.00 kJ mol^-1

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3td108

For $A B$ at $300 K$ , $Δ G^{\circ} = + 4.00 kJ mo l^{- 1}$ . Use $R = 8.314 J mo l^{- 1} K^{- 1}$ . Select the correct statement(s).

AFor $Q = 0.100$ , $Δ G \approx - 1.7 kJ mo l^{- 1}$ and the net reaction goes forward
B $K \approx 0.20$ , so $[B] / [A] \approx 0.20$ at equilibrium
CFor $Q = 1.00$ the net reaction goes forward because $B$ and $A$ are at equal concentrations
DAt equilibrium $Δ G = Δ G^{\circ} = + 4.00 kJ mo l^{- 1}$

K = exp (- 4000/ (8.314 \times 300)) = 0.201

. For

Q = 0.100

Δ G = 4000 + 2494.2 ln (0.100) = - 1.74 kJ

, negative, so forward. For

Q = 1.00

Δ G = Δ G^{\circ} = + 4.00 kJ > 0

, so the net change is backward (

Q > K

). At equilibrium

Δ G = 0

always, not

Δ G^{\circ}