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

Byju's Answer

Standard XII

Chemistry

Arrhenius Equation

The equilibri...

Question

The equilibrium constant for the reaction
${C O}_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)$ at $298 K$ is $7$ . Calculate the value of standard free energy change.
$(R = 8.314 J K^{- 1} {m o l}^{- 1})$

Open in App

Solution

We know that, $Δ G^{o} = - 2.303 R T log K_{c}$
Given, $K_{c} = 73, T = 298 K, R = 8.314 J K^{- 1} {m o l}^{- 1}$
From the above equation,
$Δ G^{o} = - 2.303 \times 8.314 \times 298 log (73) = - 10.632 k J {m o l}^{- 1}$

Suggest Corrections

Similar questions

Q. Calculate the standard internal energy change for the reaction

C_{(g r a p h i t e)} + H_{2} O (g) \to C O (g) + H_{2} O (g)

Δ_{f}^{o}

25^{o}

;

H_{2} O (g) = - 241.8 k J {m o l}^{- 1}

C O (g) = - 110.5 k J {m o l}^{- 1}

R = 8.314 J K^{- 1} {m o l}^{- 1}

Q. For the water gas reaction

C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} O (g)

the standard Gibbs free energy of reaction (at

1000 K

) is

- 8.1 k J / m o l

. Calculate its equilibrium constant.

Q. The

Δ H_{f}^{o}

for

C O_{2 (g)}

C O_{(g)}

and

H_{2} O_{(g)}

are

- 393.5

- 110.5

and

- 241.8

m o l^{- 1}

respectively the standard enthalpy change (in KJ) for the reaction

C O_{2 (g)} + H_{2 (g)} \to C O_{(g)} + H_{2} O_{(g)}

is:

Calculate the equilibrium constant ( $K_{p}$ ) for the reaction, $C (s) + C O_{2} (g) ⇌ 2 C O (g)$ , at $1300 K$ from the following data:
$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g); K_{p} (1300 K) = 3.9 a t m$

$H_{2} (g) + C O_{2} (g) ⇌ C O (g) + H_{2} O (g); K_{p} (1300 K) = 0.623 a t m$

Q. For the water gas reaction,

C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)

the standard Gibbs free energy of reaction (at

1000 K

) is

- 8.1 k J / m o l .

Calculate its equilibrium constant

Join BYJU'S Learning Program

The equilibrium constant for the reactionCO2(g)+H2(g)⇌CO(g)+H2O(g) at 298K is 7. Calculate the value of standard free energy change.(R=8.314JK−1mol−1)

We know that, ΔGo=−2.303RTlogKcGiven, Kc=73,T=298K,R=8.314JK−1mol−1From the above equation,ΔGo=−2.303×8.314×298log(73)=−10.632kJmol−1

The equilibrium constant for the reaction
${C O}_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)$ at $298 K$ is $7$ . Calculate the value of standard free energy change.
$(R = 8.314 J K^{- 1} {m o l}^{- 1})$

We know that, $Δ G^{o} = - 2.303 R T log K_{c}$
Given, $K_{c} = 73, T = 298 K, R = 8.314 J K^{- 1} {m o l}^{- 1}$
From the above equation,
$Δ G^{o} = - 2.303 \times 8.314 \times 298 log (73) = - 10.632 k J {m o l}^{- 1}$