The van't Hoff reaction isotherm is:

Δ G = {R T l o g}_{e} K_{p}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- Δ G = {R T l o g}_{e} K_{p}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Δ G = {R T}^{2} I n K_{p}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $- Δ G = {R T l o g}_{e} K_{p}$
$Δ G$
$= - R T I n K_{p}$
$= - 2.303 R T {l o g}_{10} K_{p}$

Suggest Corrections

Similar questions

Q. Assertion :At equilibrium

Δ G = 0

Reason:

Δ G^{o} = - R T {log}_{e} K_{c}

at equilibrium

Q. Which is the correct relationship between

Δ G

and equilibrium constant

K_{p}

The reaction N $_{2}$ O $_{4}$ (g) $⇌$ 2NO $_{2}$ (g) is
carried out at 298 K and 20 bar. Five moles of each of N $_{2}$ O $_{4}$ and
NO $_{2}$ , are taken initially. Given: $Δ$ $_{f}$ G $^{\circ}$
(N $_{2}$ O $_{4}$ ) = 100 kJ mol $^{- 1}$ and $Δ$ $_{f}$ G $^{\circ}$
(NO $_{2}$ ) = 50 kJ mol $^{- 1}$ .Choose the appropriate option: The values of
$Δ$ G and K $^{\circ}$ $_{p}$ at 298 K are:

Q. Which is not correct relationship between

Δ G^{⊖}

and equilibrium constant

K_{P}

C_{2} H_{6} (g) ⇌ C_{2} H_{4} (g) + H_{2} (g); K_{p} = 0.05 atm

.
The value of

Δ G^{\circ}

of the reaction at

627^{\circ} C

would be:

(Given : ln (0.05) \approx - 3; R = 8.3 {J K}^{- 1} {mol}^{- 1})

Join BYJU'S Learning Program

The van't Hoff reaction isotherm is:

The correct option is C −ΔG=RTlogeKpΔG=−RTInKp=−2.303RTlog10Kp

The correct option is C $- Δ G = {R T l o g}_{e} K_{p}$
$Δ G$
$= - R T I n K_{p}$
$= - 2.303 R T {l o g}_{10} K_{p}$