The standard free energy change of a reaction is Δ G° = - 115 kJ at 298 K. Calculate the equilibrium constant kp in log kp (R = 8.314 Jk-1 mol-1).

Question

The standard free energy change of a reaction is Δ G° = - 115 kJ at 298 K. Calculate the equilibrium constant kp in log kp (R = 8.314 Jk-1 mol-1). (A) 20.16 (B) 2.303 (C) 2.016 (D) 13.83

Tardigrade Admin · Accepted Answer

Δ G°= â 115 × 103 3 J, T = 298 K, R = â 8.314 JK-1 mol -1 â Δ G ° = 2.303RT log10K p -(- 115 × 10 3) = 2.303× 8.314 × 298 log 10 Kp log10 Kp = (115000/ 2.303 × 8.314 × 298 ) log10 K p = 20.16

Q. The standard free energy change of a reaction is $Δ G^{\circ}$ = - $115 k J$ at $298 K$ . Calculate the equilibrium constant $k_{p}$ in $lo g k_{p}$
$(R = 8.314 J k^{- 1} m o l^{- 1}) .$

20.16

2.303

2.016

13.83

$Δ G^{\circ} = - 115 \times 10 3^{3} J,$
$T = 298 K, R = - 8.314 J K^{- 1} m o l^{- 1}$
$- Δ G^{\circ} = 2.303 RT lo g_{10} K_{p}$
$- (- 115 \times 1 0^{3}) = 2.303 \times 8.314 \times 298 lo g_{10} K_{p}$
$lo g_{10} K_{p} = \frac{115000}{2.303 \times 8.314 \times 298}$
$lo g_{10} K_{p} = 20.16$

Q. The standard free energy change of a reaction is ΔG∘ = - 115kJ at 298K. Calculate the equilibrium constant kp​ in logkp​ (R=8.314Jk−1mol−1).