For the gas-phase isomerisation A B at 300 K, Delta G deg = -4.0 kJ mol^-1 (standard pressure 1 bar, R = 8.314 J mol^-1K^-1). Select all correct statements.

Starting from pure A, the equilibrium mole fraction of B is about 0.83

If Q = P_B/P_A = 10, then Delta G approx +1.74 kJ mol^-1 and the reaction goes in the reverse direction

Starting from pure A, the equilibrium mole fraction of B is about 0.83

If Q = P_B/P_A = 10, then Delta G approx +1.74 kJ mol^-1 and the reaction goes in the reverse direction

At equilibrium Delta G = Delta G deg = -4.0 kJ mol^-1

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3td219

For the gas-phase isomerisation $A B$ at $300 K$ , $Δ G^{\circ} = - 4.0 kJ mo l^{- 1}$ (standard pressure $1 bar$ , $R = 8.314 J mo l^{- 1} K^{- 1}$ ). Select all correct statements.

A $K \approx 1.60$
BStarting from pure A, the equilibrium mole fraction of B is about 0.83
CIf $Q = P_{B} / P_{A} = 10$ , then $Δ G \approx + 1.74 kJ mo l^{- 1}$ and the reaction goes in the reverse direction
DAt equilibrium $Δ G = Δ G^{\circ} = - 4.0 kJ mo l^{- 1}$

ln K = 4000/ (8.314 \times 300) = 1.604

, so

K = 4.97

; the value

1.60

ln K

, not

K

. Since

K = x_{B} / x_{A}

(equal moles),

x_{B} = K / (1 + K) = 0.833

. For

Q = 10

Δ G = - 4000 + 2494.2 \times ln 10 = + 1743 J mo l^{- 1}

, so the reverse reaction occurs. At equilibrium

Δ G = 0

, not

Δ G^{\circ}