The equilibrium constant for the reaction at 298K is-Given that- -Ho-29-8kJmol-1-So-0-100kJK-1mol-1

Given that: ΔH^o= 29.8kJmol^ 1 ΔS^o= 0.100kJK^ 1mol^ 1 T=298 K Now, ΔG^o=ΔH^o TΔS_o = 29.8 {298× ( 0.1)} Also, Δ G= RTlnK_p Δ G= RTlnK_p=0 lnK_p=0 K_p=e^0= ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Enthalpy of Combustion

The equilibri...

Question

The equilibrium constant for the reaction at $298 K$ is:
Given that: $Δ H^{o} = - 29.8 k J m o l^{- 1} Δ S^{o} = - 0.100 k J K^{- 1} m o l^{- 1}$

$1.0 \times 10^{10}$

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

$1.0 \times 10^{- 10}$

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

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

Open in App

Solution

The correct option is D 1
Given that: $Δ H^{o} = - 29.8 k J m o l^{- 1} Δ S^{o} = - 0.100 k J K^{- 1} m o l^{- 1}$
$T = 298 K$
Now, $Δ G^{o} = Δ H^{o} - T Δ S_{o}$
$= - 29.8 - {298 \times (- 0.1)}$

Also, $Δ G = - R T l n K_{p}$

$Δ G = - R T l n K_{p} = 0$

$l n K_{p} = 0$

$K_{p} = e^{0} = 1$

The equilibrium constant for the reaction at 298K is : 1

Suggest Corrections

Similar questions

Q. For the reaction,

A (g) + B (g) \to C (g) + D (g), Δ H^{o}

and

Δ S^{o}

are, respectively,

- 29.8 k J m o l^{- 1}

and

- 0.100 k J K^{- 1} m o 1^{- 1}

at 298 K.

The equilibrium constant for the reaction at 298 K is :

Q. For the reaction

2 N O C l (g) ⇌ 2 N O (g) + {C l}_{2} (g)

calculate the standard equilibrium constant at

298 K

. Given that the values of

Δ H^{o}

and

Δ S^{o}

of the reaction at

298 K

are

77.2 k J {m o l}^{- 1}

and

122 J K {m o l}^{- 1}

Q. What is the equilibrium constant

K_{c}

for the following reaction at

400 K

2 N O C l (g) ⇌ 2 N O (g) + {C l}_{2} (g)

Δ H^{o} = 77.2 k J, Δ S^{o} = 122 J K^{- 1}

400 K

Q. Variation of equilibrium constant

K

with temperature

T

is given by van't Hoff equation:

log K = log A - \frac{Δ H^{o}}{2.303 R T}

A graph between

log K

and

T^{- 1}

was a straight line as shown in the figure and having

θ = {tan}^{- 1} (- 0.5)

and

O P = 10

. Calculate:
(a)

Δ H^{o}

(Standard heat of reaction) when

T = 298 K

(b)

A

(pre-exponential factor)
(c) Equilibrium constant

K

298 K

(d)

K

798 K

, if

Δ H^{o}

is independent of temperature

Q. Assertion :Th equilibrium constant of the exothermic reaction at high temperature decreases. Reason: Since

ln \frac{K_{2}}{K_{1}} = \frac{{Δ H}^{o}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]

and for exothermic reaction,

{Δ H}^{o} = - v e

and thereby;

\frac{K_{2}}{K_{1}} < 1

Join BYJU'S Learning Program

The equilibrium constant for the reaction at 298K is: Given that: ΔHo=−29.8kJmol−1ΔSo=−0.100kJK−1mol−1

The correct option is D 1Given that: ΔHo=−29.8kJmol−1ΔSo=−0.100kJK−1mol−1 T=298K Now, ΔGo=ΔHo−TΔSo =−29.8−{298×(−0.1)} Also, ΔG=−RTlnKp ΔG=−RTlnKp=0 lnKp=0 Kp=e0=1 The equilibrium constant for the reaction at 298K is : 1

The equilibrium constant for the reaction at $298 K$ is:
Given that: $Δ H^{o} = - 29.8 k J m o l^{- 1} Δ S^{o} = - 0.100 k J K^{- 1} m o l^{- 1}$